初学推荐博客：LemonMZc BraketBN Owen_codeisking
CDQ&整体二分教程和题目：Winniechen
1

小结

CDQ分治

二维LIS

如果你还没学过CDQ分治，可以先忽略这题。。。呃

• 首先这题一看就是一个三维偏序的裸题，不过普通CDQ分治的写法会wa，因为求LIS是有后效性的。
• 我们不能先解决左边，再解决右边，最后合并，
• 比如：定义\(dp[i]\)表示以\(i\)结尾的元素的LIS的长度。
• 但是他只能记录末位置，而不知道从哪些地方转移来。
• 如果在一次合并前就解决了子区间\([mid+1,r]\)区间内的情况，这次合并就算重复的贡献。
• 因此，我们需要先解决左边，向右边转移后，再解决右边。

AC_Code

const int MXN = 1e5 + 7;
const int MXE = 1e6 + 7;
int n, m;
pii ar[MXN];
struct lp {int fi, se, val, id;
}cw[MXN];
int ans;
bool cmp(const lp&a, const lp&b) {return a.se < b.se;
}
bool cmp2(const lp&a, const lp&b) {return a.id < b.id;
}
int bit[MXN];
int N;
void bit_add(int x, int v) {while(x <= N) {bit[x] = big(bit[x], v);x += lowbit(x);}
}
void bit_clr(int x) {while(x <= N) {bit[x] = 0;x += lowbit(x);}
}
int bit_query(int x) {int ans = 0;while (x > 0) {ans = big(ans, bit[x]);x -= lowbit(x);}return ans;
}
void solve(int l, int r) {if(l >= r) return;int mid = (l + r) >> 1;solve(l, mid);sort(cw + l, cw + mid + 1, cmp);sort(cw + mid + 1, cw + r + 1, cmp);int i = l;for(int j = mid + 1; j <= r; ++j) {for(; i <= mid && cw[i].se < cw[j].se; ++i) bit_add(cw[i].fi, cw[i].val);cw[j].val = big(cw[j].val, bit_query(cw[j].fi - 1) + 1);}for(-- i; i >= l; --i) bit_clr(cw[i].fi);
//    inplace_merge(cw + l, cw + mid + 1, cw + r + 1, cmp);sort(cw + mid + 1, cw + r + 1, cmp2);solve(mid + 1, r);
}
int main() {
#ifndef ONLINE_JUDGEfreopen("/home/cwolf9/CLionProjects/ccc/in.txt", "r", stdin);
//    freopen("/home/cwolf9/CLionProjects/ccc/out.txt", "w", stdout);
#endif
//    int tim = 1;n = read();vector<int> vs;for(int i = 1; i <= n; ++i) {ar[i].fi = read(), ar[i].se = read();vs.eb(ar[i].fi);}my_unique(vs);for(int i = 1; i <= n; ++i) {ar[i].fi = lower_bound(all(vs), ar[i].fi) - vs.begin() + 1;cw[i].fi = ar[i].fi, cw[i].se = ar[i].se, cw[i].val = 1, cw[i].id = i;}N = vs.size() + 1;ans = 1;solve(1, n);for(int i = 1; i <= n; ++i) ans = big(ans, cw[i].val);printf("%d\n", ans);return 0;
}

第一道裸题 bzoj1176 Mokia

/*
2000000的矩阵，单点更新，区间查询
一维时间，一维x，树状数组维护y
*/
#include<bits/stdc++.h>
#define lson rt<<1
#define rson rt<<1|1
using namespace std;
typedef long long LL;const int MXN = 2e6 + 6;
const int INF = 0x3f3f3f3f;int n, k, s, cnt;
LL ans[MXN];
struct lp {int x, y, ip, idx, delta;
}cw[MXN];
bool cmp(const lp &a, const lp &b) {return a.x < b.x ||(a.x == b.x && a.y < b.y);
}
LL bit[MXN];
int lowbit(int x){return (x&(-x));}
void add(int x,int c){for(;x <= n;x += lowbit(x)) bit[x]+=c;}
LL query(int x){LL ans = 0;for(;x;x -= lowbit(x))ans+=bit[x];return ans;}
void cdq(int l, int r) {if(l == r) return;int mid = (l + r) >> 1;cdq(l, mid); cdq(mid+1,r);int j = l;for(int i = mid+1; i <= r; ++i) {if(cw[i].ip == 2) {for(; j <= mid && cw[j].x <= cw[i].x; ++j) {if(cw[j].ip == 1) add(cw[j].y, cw[j].delta);}ans[cw[i].idx] += cw[i].delta * query(cw[i].y);}}for(-- j; j >= l; --j) if(cw[j].ip == 1) add(cw[j].y, -cw[j].delta);inplace_merge(cw+l,cw+mid+1,cw+r+1, cmp);
}
int main() {scanf("%d%d", &s, &n);cnt = 0;int cntp = 0, x, y, a, b;while(scanf("%d", &k) == 1 && k != 3) {if(k == 1) {scanf("%d%d%d", &x, &y, &a);++ cnt;cw[cnt] = {x, y, 1, cnt, a};}else {scanf("%d%d%d%d", &x, &y, &a, &b);++ cntp; ans[cntp] = (1LL*a - x)*(b - y)*s;++ cnt; cw[cnt] = {x-1, y-1, 2, cntp, 1};++ cnt; cw[cnt] = {x-1, b, 2, cntp, -1};++ cnt; cw[cnt] = {a, y-1, 2, cntp, -1};++ cnt; cw[cnt] = {a, b, 2, cntp, 1};}}cdq(1, cnt);for(int i = 1; i <= cntp; ++i) printf("%lld\n", ans[i]);return 0;
}

bzoj3262 陌上花开

/*
简单地说，对于每个点求出x,y,z都小于等于它的点的个数
一维x，一维y，树状数组维护z
*/
#include<bits/stdc++.h>
#define lson rt<<1
#define rson rt<<1|1
using namespace std;
typedef long long LL;const int MXN = 2e6 + 6;
const int INF = 0x3f3f3f3f;int n, s;
struct lp {int x, y, z, ip, cnt, ans;
}cw[MXN], temp[MXN];
bool cmpx(const lp &a, const lp &b) {return a.x < b.x ||(a.x == b.x && a.y < b.y)||(a.x == b.x && a.y == b.y && a.z < b.z);
}
bool cmpyz(const lp &a, const lp &b) {return a.y < b.y||(a.y == b.y && a.z < b.z);
}
int bit[MXN];
int lowbit(int x){return (x&(-x));}
void add(int x,int c){for(;x <= s;x += lowbit(x)) bit[x]+=c;}
int query(int x){int ans = 0;for(;x;x -= lowbit(x))ans+=bit[x];return ans;}
void cdq(int l,int r) {if(l == r) return ;int mid = (l + r) >> 1;cdq(l, mid); cdq(mid+1, r);//sort(cw + l, cw + 1 + mid, cmpyz);   //sort(cw + 1 + mid, cw + 1 + r, cmpyz);int j = l;for(int i = mid+1; i <= r; ++i) {for(; j <= mid && cw[j].y <= cw[i].y; ++j) {add(cw[j].z, cw[j].cnt);}cw[i].ans += query(cw[i].z);}for(-- j; j >= l; --j) add(cw[j].z, -cw[j].cnt);int p1 = l, p2 = mid+1, cnt = l;while(p1 <= mid && p2 <= r) {if(p1 <= mid && cw[p1].y <= cw[p2].y) temp[cnt++] = cw[p1], ++ p1;else temp[cnt++] = cw[p2], ++ p2;}while(p1 <= mid) temp[cnt++] = cw[p1], ++ p1;while(p2 <= r) temp[cnt++] = cw[p2], ++ p2;for(int i = l; i <= r; ++i) cw[i] = temp[i];
}
int main() {scanf("%d%d", &n, &s);for(int i = 1; i <= n; ++i) {scanf("%d%d%d", &cw[i].x, &cw[i].y, &cw[i].z);cw[i].ip = i; cw[i].ans = 0;}sort(cw + 1, cw + 1 + n, cmpx);int cnt = 1, p = 0;for(int i = 2; i <= n; ++i) {if(cw[i].x!=cw[i-1].x||cw[i].y!=cw[i-1].y||cw[i].z!=cw[i-1].z) {cw[++p] = cw[i-1];cw[p].cnt = cnt;cnt = 1;if(i == n) {cw[++p] = cw[i];cw[p].cnt = cnt;}}else {++ cnt;if(i == n) {cw[++p] = cw[i-1];cw[p].cnt = cnt;}}}cdq(1, p);for(int i = 1; i <= p; ++i) bit[cw[i].ans+cw[i].cnt-1] += cw[i].cnt;for(int i = 0; i < n; ++i) printf("%d\n", bit[i]);return 0;
}

bzoj 1790 矩形藏宝地

/*
对每个点询问是否存在一个点a,b小于等于它，c,d大于等于它
一维c，一维d，离散化a，线段树维护a上的b
*/
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;const int INF = 0x3f3f3f3f;
const int MXN = 1e6 + 6;int n, m, k, ans;
struct lp {int a, b, c, d, ans;
}cw[MXN], temp[MXN];
int ar[MXN], sum[MXN<<2];
bool cmpc(const lp &A, const lp &B) {return A.c > B.c;
}
bool cmpd(const lp &A, const lp &B) {return A.d > B.d;
}
void update(int p,int val,int l,int r,int rt) {if(l == r) {sum[rt] = val;return ;}int mid = (l + r) >> 1;if(p <= mid) update(p, val, l, mid, rt<<1);else update(p, val, mid+1, r, rt<<1|1);sum[rt] = min(sum[rt<<1], sum[rt<<1|1]);
}
int query(int L,int R,int l,int r,int rt) {if(L <= l && r <= R) {return sum[rt];}int mid = (l + r) >> 1;if(L > mid) return query(L,R,mid+1,r,rt<<1|1);else if(R <= mid) return query(L,R,l,mid,rt<<1);else {return min(query(L,mid,l,mid,rt<<1),query(mid+1,R,mid+1,r,rt<<1|1));}
}
void cdq(int l,int r) {if(l == r) return ;int mid = (l + r) >> 1;cdq(l, mid); cdq(mid+1, r);int j = l;for(int i = mid+1; i <= r; ++i) {for(; j <= mid && cw[j].d >= cw[i].d; ++j) {update(cw[j].a, cw[j].b, 1, k, 1);}if(query(1,cw[i].a,1,k,1) <= cw[i].b) cw[i].ans = 1;}for(-- j; j >= l; --j) update(cw[j].a, INF, 1, k, 1);int p1 = l, p2 = mid+1, cnt = l;while(p1 <= mid && p2 <= r) {if(p1 <= mid && cw[p1].d >= cw[p2].d) temp[cnt++] = cw[p1], ++ p1;else temp[cnt++] = cw[p2], ++ p2;}while(p1 <= mid) temp[cnt++] = cw[p1], ++ p1;while(p2 <= r) temp[cnt++] = cw[p2], ++ p2;for(int i = l; i <= r; ++i) cw[i] = temp[i];
}
int main(){scanf("%d", &n);for(int i = 1; i <= n; ++i) {scanf("%d%d%d%d", &cw[i].a, &cw[i].b, &cw[i].c, &cw[i].d);ar[i] = cw[i].a; cw[i].ans = 0;}memset(sum, 0x3f, sizeof(sum));sort(ar+1, ar+1+n);k = unique(ar+1,ar+1+n) - ar;sort(cw + 1, cw + 1 + n, cmpc);for(int i = 1; i <= n; ++i) {cw[i].a = lower_bound(ar+1,ar+k,cw[i].a) - ar;cw[i].c = i;}cdq(1, n);for(int i = 1; i <= n; ++i) ans += cw[i].ans;printf("%d\n", ans);return 0;
}

hdu5126四维偏序

/*
三维空间n次操作，更新一个点，查询两点的长方体间点的个数
CDQ套CDQ：一维时间，一维x，再一维y，树状数组维护z
每次查询，三维容斥一下，分成8个区间的查询
*/
#include<bits/stdc++.h>
#define lson rt<<1
#define rson rt<<1|1
using namespace std;
typedef long long LL;const int MXN = 2e6 + 6;
const int INF = 0x3f3f3f3f;int n, tot, k;
struct lp {int x, y, z, flag, idx, ip, delta;
}cw[MXN], temp[MXN], Tmp[MXN];
int ans[MXN], ar[MXN];
int bit[MXN];
int lowbit(int x){return (x&(-x));}
void add(int x,int c){for(;x <= 400006;x += lowbit(x)) bit[x]+=c;}
int query(int x){int ANs = 0;for(;x;x -= lowbit(x))ANs+=bit[x];return ANs;}
void cdq2(int l,int r) {if(l == r) return;int mid = (l + r) >> 1;cdq2(l, mid); cdq2(mid+1, r);int j = l;for(int i = mid+1; i <= r; ++i) {for(; j <= mid && temp[j].y <= temp[i].y; ++j)if(temp[j].ip == 1&&temp[j].flag == 0) add(temp[j].z, 1);if(temp[i].ip == 2&&temp[i].flag == 1) {ans[temp[i].idx] += query(temp[i].z)*temp[i].delta;}}for(-- j; j >= l; --j) if(temp[j].ip == 1&&temp[j].flag == 0) add(temp[j].z, -1);int p1 = l, p2 = mid+1, cnt = l;while(p1 <= mid && p2 <= r) {if(temp[p1].y <= temp[p2].y) {Tmp[cnt++] = temp[p1]; ++ p1;}else {Tmp[cnt++] = temp[p2]; ++ p2;}}while(p1 <= mid) {Tmp[cnt++] = temp[p1]; ++ p1;}while(p2 <= r) {Tmp[cnt++] = temp[p2]; ++ p2;}for(int i = l; i <= r; ++i) temp[i] = Tmp[i];
}
void cdq1(int l,int r) {if(l == r) return ;int mid = (l + r) >> 1;cdq1(l, mid); cdq1(mid+1, r);int p1 = l, p2 = mid+1, cnt = l;while(p1 <= mid && p2 <= r) {if(cw[p1].x <= cw[p2].x) {temp[cnt++] = cw[p1]; ++ p1;temp[cnt-1].flag = 0;}else {temp[cnt++] = cw[p2]; ++ p2;temp[cnt-1].flag = 1;}}while(p1 <= mid) {temp[cnt++] = cw[p1]; ++ p1;temp[cnt-1].flag = 0;}while(p2 <= r) {temp[cnt++] = cw[p2]; ++ p2;temp[cnt-1].flag = 1;}for(int i = l; i <= r; ++i) cw[i] = temp[i];cdq2(l, r);
}
int main() {int tim; scanf("%d", &tim);while(tim --) {scanf("%d", &n);tot = 0;int totp = 0;for(int i = 0; i < n; ++i) {int opt, x1, y1, z1, x2, y2, z2;scanf("%d%d%d%d", &opt, &x1, &y1, &z1);if(opt == 1) {++ tot; cw[tot] = {x1, y1, z1, 0, 0, 1, 0};}else {scanf("%d%d%d", &x2, &y2, &z2);++ totp;++ tot; cw[tot] = {x2, y2, z2, 0, totp, 2, 1};++ tot; cw[tot] = {x1-1, y2, z2, 0, totp, 2, -1};++ tot; cw[tot] = {x2, y1-1, z2, 0, totp, 2, -1};++ tot; cw[tot] = {x2, y2, z1-1, 0, totp, 2, -1};++ tot; cw[tot] = {x1-1, y1-1, z2, 0, totp, 2, 1};++ tot; cw[tot] = {x1-1, y2, z1-1, 0, totp, 2, 1};++ tot; cw[tot] = {x2, y1-1, z1-1, 0, totp, 2, 1};++ tot; cw[tot] = {x1-1, y1-1, z1-1, 0, totp, 2, -1};}}for(int i = 1; i <= tot; ++i) ar[i] = cw[i].z;sort(ar+1, ar+1+tot);k = unique(ar+1, ar+1+tot) - ar;for(int i = 1; i <= tot; ++i) cw[i].z = lower_bound(ar+1, ar+k, cw[i].z)-ar;cdq1(1, tot);for(int i = 1; i <= totp; ++i) printf("%d\n", ans[i]), ans[i] = 0;}return 0;
}

P3157 [CQOI2011]动态逆序对

CDQ

/*
给1到n的一个排列，按某顺序依次删除m个元素，求在每次删除一个元素之前序列的逆序对数。
*/
//对不起，我太暴力了
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;const int MXN = 1e5 + 6;
const int INF = 0x3f3f3f3f;int n, m;
LL tot;
struct lp {int x, tim, pos;LL ans;
}cw[MXN], temp[MXN], Tmp[MXN];
LL ans[MXN], ar[MXN];
LL bit[MXN];
int lowbit(int x){return (x&(-x));}
void ADD(int x,int c){for(; x <= n; x += lowbit(x)) bit[x]+=c;}
LL QUERY(int x){LL ANS = 0;for(; x; x -= lowbit(x)) ANS+=bit[x];return ANS;}
bool cmppos(const lp &a, const lp &b) {return a.pos > b.pos;
}
void cdq1(int l,int r) {//求出每个数左边有多少个时间比你大的且值比你大的数if(l == r) return ;int mid = (l + r) >> 1;cdq1(l, mid); cdq1(mid+1, r);int j = l;for(int i = mid + 1; i <= r; ++i) {for(; j <= mid && cw[j].tim >= cw[i].tim; ++j) {if(cw[j].tim) ADD(cw[j].x, 1);}cw[i].ans += QUERY(n) - QUERY(cw[i].x);}for(-- j; j >= l; --j) if(cw[j].tim) ADD(cw[j].x, -1);int p1 = l, p2 = mid+1, cnt = l;while(p1 <= mid && p2 <= r) {if(cw[p1].tim >= cw[p2].tim) temp[cnt++] = cw[p1], ++p1;else temp[cnt++] = cw[p2], ++ p2;}while(p1 <= mid) temp[cnt++] = cw[p1], ++p1;while(p2 <= r) temp[cnt++] = cw[p2], ++ p2;for(int i = l; i <= r; ++i) cw[i] = temp[i];
}
void cdq2(int l,int r) {//求出每个数右边有多少个时间比你大的且值比你小的数if(l == r) return ;int mid = (l + r) >> 1;cdq2(l, mid); cdq2(mid+1, r);int j = l;for(int i = mid + 1; i <= r; ++i) {for(; j <= mid && cw[j].tim >= cw[i].tim; ++j) {if(cw[j].tim) ADD(cw[j].x, 1);}cw[i].ans += QUERY(cw[i].x);}for(-- j; j >= l; --j) if(cw[j].tim) ADD(cw[j].x, -1);int p1 = l, p2 = mid+1, cnt = l;while(p1 <= mid && p2 <= r) {if(cw[p1].tim >= cw[p2].tim) temp[cnt++] = cw[p1], ++p1;else temp[cnt++] = cw[p2], ++ p2;}while(p1 <= mid) temp[cnt++] = cw[p1], ++p1;while(p2 <= r) temp[cnt++] = cw[p2], ++ p2;for(int i = l; i <= r; ++i) cw[i] = temp[i];
}
void cdq(int l,int r) {if(l == r) return ;int mid = (l + r) >> 1;cdq(l, mid); cdq(mid+1, r);int p1 = l, p2 = mid+1, cnt = l;while(p1 <= mid && p2 <= r) {if(cw[p1].x <= cw[p2].x) temp[cnt++] = cw[p1], ++p1;else temp[cnt++] = cw[p2], ++ p2, tot += mid-p1+1;}while(p1 <= mid) temp[cnt++] = cw[p1], ++p1;while(p2 <= r) temp[cnt++] = cw[p2], ++ p2;for(int i = l; i <= r; ++i) cw[i] = temp[i];
}
int main() {scanf("%d%d", &n, &m);for(int i = 1; i <= n; ++i) {scanf("%d", &cw[i].x);cw[i].tim = 0;cw[i].ans = 0;ar[cw[i].x] = i;cw[i].pos = i;}for(int i = 1, x; i <= m; ++i) {scanf("%d", &x);cw[ar[x]].tim = i;}cdq(1, n);sort(cw + 1, cw + 1 + n, cmppos);reverse(cw + 1, cw + 1 + n);for(int i = 1; i <= n; ++i) {if(cw[i].tim == 0) {ADD(cw[i].x, 1);}else {ans[cw[i].tim] += QUERY(n) - QUERY(cw[i].x);}}memset(bit, 0, sizeof(LL)*(n+10));for(int i = n; i >= 1; --i) {if(cw[i].tim == 0) {ADD(cw[i].x, 1);}else {ans[cw[i].tim] += QUERY(cw[i].x);}}memset(bit, 0, sizeof(LL)*(n+10));cdq1(1, n);for(int i = 1; i <= n; ++i) {ans[cw[i].tim] += cw[i].ans;cw[i].ans = 0;}sort(cw + 1, cw + 1 + n, cmppos);cdq2(1, n);for(int i = 1; i <= n; ++i) {ans[cw[i].tim] += cw[i].ans;}printf("%lld\n", tot);for(int i = 1; i < m; ++i) {tot -= ans[i];printf("%lld\n", tot);}return 0;
}

树套树

/*
树状数组套权值线段树
这东西是干什么的呢？
个人感觉大概就是把那种O(n)*O(1)的复杂度优化成O(log)*O(log)的复杂度
比如n个位置，每个位置一个值，查询是O(1)的；log个位置(树状数组)，查询是O(log)的(权值线段树/主席树)
本题把每次询问拆分成两个前缀和相减，然后用树套树优化，树状数组每个节点维护的范围不变，然后将节点变成权值线段树，就是本题的做法。
*/
#include<bits/stdc++.h>
#define lson rt<<1
#define rson rt<<1|1
using namespace std;
typedef long long LL;const int MXN = 2e5 + 6;
const int MXE = 6e6 + 6;
const int INF = 0x3f3f3f3f;int n, m;
struct lp {int l, r;LL sum;
}cw[MXE], temp[MXN], Tmp[MXN];
int lb[MXN], rs[MXN], ar[MXN], bel[MXN];
int Root[MXN], NODE, A[MXE], B[MXE];
LL bit[MXN];
int lowbit(int x){return (x&(-x));}
void ADD(int x,int c){for(; x <= n; x += lowbit(x)) bit[x]+=c;}
LL QUERY(int x){LL ANS = 0;for(; x; x -= lowbit(x)) ANS+=bit[x];return ANS;}bool cmppos(const lp &a, const lp &b) {return a.sum > b.sum;
}
void update(int &cur, int x, int l, int r) {if(!cur) {cur = ++ NODE;cw[cur].l = cw[cur].r = 0;}++ cw[cur].sum;if(l == r) return ;int mid = (l + r) >> 1;if(x <= mid) update(cw[cur].l, x, l, mid);else update(cw[cur].r, x, mid+1, r);
}
LL Find(int l,int r,int lw,int up,int rt) {if(lw <= l && r <= up) {return cw[rt].sum;}if(lw > r || up < l) return 0;int mid = (l + r) >> 1;if(lw > mid) return Find(mid+1, r, lw, up, cw[rt].r);else if(up <= mid) return Find(l, mid, lw, up, cw[rt].l);return Find(l, mid, lw, mid, cw[rt].l)+Find(mid+1,r,mid+1,up,cw[rt].r);
}
LL query(int l,int r,int lw,int up) {if(lw > up) return 0;if(l >= r) return 0;A[0] = B[0] = 0;for(; r; r -= lowbit(r)) A[++A[0]] = Root[r];//树状数组的求和，每个节点都是一个权值线段树，每个节点的查询时O(log)的for(; l; l -= lowbit(l)) B[++B[0]] = Root[l];LL ans = 0;for(int i = 1; i <= A[0]; ++i) ans += Find(1,n,lw,up,A[i]);for(int i = 1; i <= B[0]; ++i) ans -= Find(1,n,lw,up,B[i]);return ans;
}
int main() {scanf("%d%d", &n, &m);for(int i = 1; i <= n; ++i) {scanf("%d", &ar[i]);bel[ar[i]] = i;}LL tot = 0;for(int i = 1; i <= n; ++i) {lb[i] = QUERY(n) - QUERY(ar[i]);ADD(ar[i], 1);tot += lb[i];cw[i].l = cw[i].r = 0;}memset(bit, 0, sizeof(LL)*(n+5));for(int i = n; i >= 1; --i) {rs[i] = QUERY(ar[i]-1);ADD(ar[i], 1);}memset(bit, 0, sizeof(LL)*(n+5));printf("%lld\n", tot);NODE = n;for(int i = 1, x; i < m; ++i) {scanf("%d", &x);tot -= (lb[bel[x]] + rs[bel[x]] - query(0,bel[x]-1,x+1,n) - query(bel[x],n,1,x-1));printf("%lld\n", tot);for(int y = bel[x]; y <= n; y += lowbit(y)) {//将所有包含bel[x]位置的权值线段树更新update(Root[y], x, 1, n);}}scanf("%d", &m);return 0;
}
//这样写跑的快一点
#include<bits/stdc++.h>
#define lson rt<<1
#define rson rt<<1|1
using namespace std;
typedef long long LL;const int MXN = 2e5 + 6;
const int MXE = 6e6 + 6;
const int INF = 0x3f3f3f3f;int n, m;
struct lp {int l, r;LL sum;
}cw[MXE], temp[MXN], Tmp[MXN];
int lb[MXN], rs[MXN], ar[MXN], bel[MXN];
int Root[MXN], NODE, A[MXE], B[MXE];
LL bit[MXN];
int lowbit(int x){return (x&(-x));}
void ADD(int x,int c){for(; x <= n; x += lowbit(x)) bit[x]+=c;}
LL QUERY(int x){LL ANS = 0;for(; x; x -= lowbit(x)) ANS+=bit[x];return ANS;}bool cmppos(const lp &a, const lp &b) {return a.sum > b.sum;
}
void update(int &cur, int x, int l, int r) {if(!cur) {cur = ++ NODE;cw[cur].l = cw[cur].r = 0;}++ cw[cur].sum;if(l == r) return ;int mid = (l + r) >> 1;if(x <= mid) update(cw[cur].l, x, l, mid);else update(cw[cur].r, x, mid+1, r);
}
LL queryda(int l,int r,int x) {if(l >= r) return 0;A[0] = B[0] = 0;for(; r; r -= lowbit(r)) A[++A[0]] = Root[r]; for(; l; l -= lowbit(l)) B[++B[0]] = Root[l];LL ans = 0;l = 1, r = n;int mid;while(l + 1 <= r) {mid = (l + r) >> 1;if(x <= mid) {for(int i = 1; i <= A[0]; ++i) ans += cw[cw[A[i]].r].sum;for(int i = 1; i <= B[0]; ++i) ans -= cw[cw[B[i]].r].sum;for(int i = 1; i <= A[0]; ++i) A[i] = cw[A[i]].l;for(int i = 1; i <= B[0]; ++i) B[i] = cw[B[i]].l;r = mid;}else {for(int i = 1; i <= A[0]; ++i) A[i] = cw[A[i]].r;for(int i = 1; i <= B[0]; ++i) B[i] = cw[B[i]].r;l = mid + 1;}}return ans;
}
LL queryxiao(int l,int r,int x) {if(l >= r) return 0;A[0] = B[0] = 0;for(; r; r -= lowbit(r)) A[++A[0]] = Root[r]; for(; l; l -= lowbit(l)) B[++B[0]] = Root[l];LL ans = 0;l = 1, r = n;int mid;x - 1;while(l + 1 <= r) {mid = (l + r) >> 1;if(x >= mid + 1) {for(int i = 1; i <= A[0]; ++i) ans += cw[cw[A[i]].l].sum;for(int i = 1; i <= B[0]; ++i) ans -= cw[cw[B[i]].l].sum;for(int i = 1; i <= A[0]; ++i) A[i] = cw[A[i]].r;for(int i = 1; i <= B[0]; ++i) B[i] = cw[B[i]].r;l = mid + 1;}else {for(int i = 1; i <= A[0]; ++i) A[i] = cw[A[i]].l;for(int i = 1; i <= B[0]; ++i) B[i] = cw[B[i]].l;r = mid;}}return ans;
}
int main() {scanf("%d%d", &n, &m);for(int i = 1; i <= n; ++i) {scanf("%d", &ar[i]);bel[ar[i]] = i;}LL tot = 0;for(int i = 1; i <= n; ++i) {lb[i] = QUERY(n) - QUERY(ar[i]);ADD(ar[i], 1);tot += lb[i];cw[i].l = cw[i].r = 0;}memset(bit, 0, sizeof(LL)*(n+5));for(int i = n; i >= 1; --i) {rs[i] = QUERY(ar[i]-1);ADD(ar[i], 1);}memset(bit, 0, sizeof(LL)*(n+5));printf("%lld\n", tot);NODE = n;for(int i = 1, x; i < m; ++i) {scanf("%d", &x);tot -= (lb[bel[x]]+rs[bel[x]]-queryda(0,bel[x]-1,x)-queryxiao(bel[x],n,x));printf("%lld\n", tot);for(int y = bel[x]; y <= n; y += lowbit(y)) {update(Root[y], x, 1, n);}}scanf("%d", &m);return 0;
}

bzoj1176 树套树但是我过不了

我不会平衡树昂，就不能让我套个线段树吗？你疯狂给我mle&re是个啥意思？难度要去学平衡树吗？。。不，我不学。。。难受啊

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MXN = 2e5 + 5;
const int MXE = 1e7 + 3e6 + 6;
const int INF = 0x3f3f3f3f;
int n, m;
int s, p, k;
int Root[MXN], NODE;
int ar[MXN], br[MXN];
struct lp {int x, y, X, Y, idx;
}cw[MXN];
struct tree {int l, r, sum;
}edge[MXE];
int lowbit(int x){return (x&(-x));}
void update(int &p, int pos, int x, int l, int r) {if(!p) {p = ++ NODE;edge[p].l = edge[p].r = edge[p].sum = 0;}edge[p].sum += x;if(l == r) return;int mid = (l + r) >> 1;if(pos <= mid) update(edge[p].l, pos, x, l, mid);else update(edge[p].r, pos, x, mid+1, r);
}
int query(int L, int R, int l, int r, int rt) {if(L <= l && r <= R) {return edge[rt].sum;}if(L > r || R < l) return 0;int mid = (l + r) >> 1;if(L > mid) return query(L, R, mid + 1, r, edge[rt].r);else if(R <= mid) return query(L, R, l, mid, edge[rt].l);return query(L,mid,l,mid,edge[rt].l)+query(mid+1,R,mid+1,r,edge[rt].r);
}
int CALC(int X, int Y) {int ans = 0;if(X == 0 || Y == 0) return ans;for(int u = X; u; u -= lowbit(u)) {ans += query(1, Y, 1, p, Root[u]);}return ans;
}
int main() {scanf("%d%d", &s, &m);int tot;for(int i = 1;; ++ i) {tot = i;scanf("%d", &cw[i].idx);if(cw[i].idx == 1) {scanf("%d%d", &cw[i].x, &cw[i].y);scanf("%d", &cw[i].X);}else if(cw[i].idx == 2) {scanf("%d%d", &cw[i].x, &cw[i].y);scanf("%d%d", &cw[i].X, &cw[i].Y);}else break;}-- tot;k = 0, p = 0;for(int i = 1; i <= tot; ++i) {ar[++ k] = cw[i].x;if(cw[i].idx == 2) ar[++ k] = cw[i].X;br[++ p] = cw[i].y;if(cw[i].idx == 2) br[++ p] = cw[i].Y;}sort(ar + 1, ar + k + 1);sort(br + 1, br + p + 1);k = unique(ar + 1, ar + k + 1) - ar;p = unique(br + 1, br + p + 1) - br;NODE = p-1;for(int i = 1; i <= tot; ++i) {cw[i].x = lower_bound(ar + 1, ar + k, cw[i].x)-ar;cw[i].y = lower_bound(br + 1, br + p, cw[i].y)-br;if(cw[i].idx == 1) {for(int y = cw[i].x; y <= k; y += lowbit(y)) {update(Root[y], cw[i].y, cw[i].X, 1, p);}}else {cw[i].X = lower_bound(ar + 1, ar + k, cw[i].X)-ar;cw[i].Y = lower_bound(br + 1, br + p, cw[i].Y)-br;printf("%d\n", CALC(cw[i].X, cw[i].Y)-CALC(cw[i].X,cw[i].y-1)-CALC(cw[i].x-1,cw[i].Y)+CALC(cw[i].x-1,cw[i].y-1));}}return 0;
}

CF 762E

CDQ写法

/*
转化为二维单点更新区间求和
*/
#include<bits/stdc++.h>
#define lson rt<<1
#define rson rt<<1|1
using namespace std;
typedef long long LL;const int MXN = 2e6 + 5;
const int MXE = 1e7 + 3e6 + 6;
const int INF = 0x3f3f3f3f;int n, m, N, k, cnt;
LL Ans;
struct lh {int x, r, f;
}edge[MXN];
struct lp {int flag, x, y, idx, delta;
}cw[MXN], temp[MXN];
int ar[MXN], ans[MXN];
bool cmp(const lh &a, const lh &b) {return a.r > b.r;
}
int bit[MXN];
int lowbit(int x){return (x&(-x));}
void ADD(int x,int c,int N){for(; x <= N; x += lowbit(x)) bit[x]+=c;}
int QUERY(int x){if(x<=0)return 0;int ANS = 0;for(; x; x -= lowbit(x)) ANS+=bit[x];return ANS;}void solve(int l, int r) {if(l == r) return ;int mid = (l + r) >> 1;solve(l, mid); solve(mid + 1, r);int j = l;for(int i = mid + 1; i <= r; ++i) {for(; j <= mid && cw[j].x <= cw[i].x; ++j) {if(cw[j].flag == 0) ADD(cw[j].y, 1, N);}if(cw[i].flag) Ans += QUERY(cw[i].y)*cw[i].delta;}for(-- j; j >= l; -- j) if(cw[j].flag == 0) ADD(cw[j].y, -1, N);int p1 = l, p2 = mid + 1, CNT = l;while(p1 <= mid && p2 <= r) {if(cw[p1].x <= cw[p2].x) temp[CNT++] = cw[p1], ++p1;else temp[CNT++] = cw[p2], ++p2;}while(p1 <= mid) temp[CNT++] = cw[p1], ++p1;while(p2 <= r) temp[CNT++] = cw[p2], ++p2;for(int i = l; i <= r; ++i) cw[i] = temp[i];
}
int main() {scanf("%d%d", &n, &m);for(int i = 1; i <= n; ++i) {scanf("%d%d%d", &edge[i].x, &edge[i].r, &edge[i].f);++ edge[i].f; N = max(N, edge[i].f + m + 1);}sort(edge + 1, edge + 1 + n, cmp);for(int i = 1, a, b, c, d; i <= n; ++i) {a = edge[i].x-edge[i].r;b = edge[i].x+edge[i].r;c = max(edge[i].f - m, 1); d = edge[i].f + m;//(a, c), (b, d)++cnt; cw[cnt] = {1, b, d, i, 1};++cnt; cw[cnt] = {1, a-1, d, i, -1};++cnt; cw[cnt] = {1, b, c-1, i, -1};++cnt; cw[cnt] = {1, a-1, c-1, i, 1};++cnt; cw[cnt] = {0, edge[i].x, edge[i].f, i, 0};}solve(1, cnt);printf("%lld\n", Ans);return 0;
}

树套树写法

/*
由于本题f的变化范围只有10，所以建普通权值线段树，然后暴力查询21次就行。
但是如果f的变化范围很大的话，可能必须得cdq分治或树套树才能写，下面是树套树代码，原理和动态逆序对那题一样。
*/
#include<bits/stdc++.h>
#define lson rt<<1
#define rson rt<<1|1
using namespace std;
typedef long long LL;const int MXN = 5e5 + 5;
const int MXE = 1e7 + 6;
const int INF = 0x3f3f3f3f;int n, m, k, N;
LL Ans;
struct lh {int x, r, f;
}edg[MXN];
struct lp {int l, r, sum;
}cw[MXE];
int Root[MXN], NODE;
int ar[MXN], ans[MXN];
bool cmp(const lh &a, const lh &b) {return a.r > b.r;
}
int bit[MXN];
int lowbit(int x){return (x&(-x));}
void ADD(int x,int c,int N){for(; x <= N; x += lowbit(x)) bit[x]+=c;}
int QUERY(int x){if(x<=0)return 0;int ANS = 0;for(; x; x -= lowbit(x)) ANS+=bit[x];return ANS;}void update(int &p, int x, int l, int r) {if(!p) p = ++ NODE;++ cw[p].sum;if(l == r) return;int mid = (l + r) >> 1;if(x <= mid) update(cw[p].l, x, l, mid);else update(cw[p].r, x, mid+1, r);
}
int query(int rt, int L, int R, int l, int r) {if(rt == 0) return 0;if(L <= l && r <= R) {return cw[rt].sum;}if(L > r || R < l) return 0;int mid = (l + r) >> 1;if(L > mid) return query(cw[rt].r, L, R, mid+1, r);else if(R <= mid) return query(cw[rt].l, L, R, l, mid);else {return query(cw[rt].l, L, mid, l, mid) + query(cw[rt].r, mid+1, R, mid + 1, r);}
}
int calc(int a, int b, int l, int r) {int ans = 0;for(; a; a -= lowbit(a)) ans -= query(Root[a], l, r, 1, k);for(; b; b -= lowbit(b)) ans += query(Root[b], l, r, 1, k);return ans;
}
int main() {scanf("%d%d", &n, &m);for(int i = 1; i <= n; ++i) {scanf("%d%d%d", &edg[i].x, &edg[i].r, &edg[i].f);ar[++k] = edg[i].x; ar[++k] = edg[i].x-edg[i].r;ar[++k] = edg[i].x+edg[i].r; N = max(N, edg[i].f + m);}sort(edg + 1, edg + 1 + n, cmp);sort(ar + 1, ar + 1 + k);k = unique(ar + 1, ar + 1 + k) - ar;for(int i = 1, l, r; i <= n; ++i) {l = lower_bound(ar + 1, ar + k, edg[i].x-edg[i].r) - ar;r = lower_bound(ar + 1, ar + k, edg[i].x+edg[i].r) - ar;Ans += calc(max(1, edg[i].f-m)-1, edg[i].f + m, l, r);l = lower_bound(ar + 1, ar + k, edg[i].x) - ar;for(int u = edg[i].f; u <= N; u += lowbit(u)) {update(Root[u], l, 1, k);}}printf("%lld\n", Ans);return 0;
}

CSUSTOJ 1024：CDQ

#include<bits/stdc++.h>
#define eb emplace_back
#define all(x) (x).begin(), (x).end()
using namespace std;
typedef long long LL;const int INF = 0x3f3f3f3f;
const int MXN = 3e6 + 6;int n, m;
struct lp {int ip, x, p, delta, id;
}cw[MXN], temp[MXN];
struct node {int opt, x, y, old;
}Fk[MXN];
int ALL;
int ar[MXN], ANS[MXN], QWE[MXN], sum[MXN<<2], br[MXN], is[MXN];
int bit[MXN];
int lowbit(int x){return (x&(-x));}
void bit_add(int x,int c){for(;x <= n+1;x += lowbit(x)) bit[x]^=c;}
int bit_query(int x){int ans = 0;for(;x;x -= lowbit(x))ans^=bit[x];return ans;}
void update(int p,int val,int l,int r,int rt) {is[p] += val;return;
}
int query(int p,int l,int r,int rt) {return is[p];
}
void cdq(int l, int r) {if(l == r) return ;//printf("[%d, %d]\n", l, r);int mid = (l + r) >> 1;cdq(l, mid); cdq(mid+1, r);int j = l;for(int i = mid+1; i <= r; ++i) {for(; j <= mid && cw[j].p <= cw[i].p; ++j) {if(cw[j].ip == 1) {if(cw[j].id != -1) update(cw[j].id, -1, 1, ALL, 1);update(cw[j].x, 1, 1, ALL, 1);}}if(cw[i].ip == 2) {ANS[cw[i].id] += cw[i].delta*query(cw[i].x, 1, ALL, 1);//printf("%d %d %d %d*\n", cw[i].id, cw[i].delta*query(cw[i].x, 1, ALL, 1), cw[i].x, cw[i].delta);}}for(-- j; j >= l; --j) if(cw[j].ip == 1) {if(cw[j].id != -1) update(cw[j].id, 1, 1, ALL, 1);update(cw[j].x, -1, 1, ALL, 1);}int p1 = l, p2 = mid+1, cnt = l;while(p1 <= mid && p2 <= r) {if(p1 <= mid && cw[p1].p <= cw[p2].p) temp[cnt++] = cw[p1], ++ p1;else temp[cnt++] = cw[p2], ++ p2;}while(p1 <= mid) temp[cnt++] = cw[p1], ++ p1;while(p2 <= r) temp[cnt++] = cw[p2], ++ p2;for(int i = l; i <= r; ++i) cw[i] = temp[i];
}
int main(){
#ifndef ONLINE_JUDGEfreopen("E://ADpan//in.in", "r", stdin);freopen("E://ADpan//out.out", "w", stdout);
#endif//printf("%d\n", (int)(200000*log(1000000)*log(1000000)));std::vector<int> vs;vs.eb(-1);scanf("%d%d", &n, &m);assert(n >= 1); assert(n <= 1000000);assert(m >= 1); assert(m <= 100000);for(int i = 1; i <= n; ++i) {scanf("%d", &ar[i]), bit_add(i, ar[i]), vs.eb(ar[i]), br[i] = ar[i];assert(ar[i] >= 0); assert(ar[i] <= 1000000000);}for(int i = 1, x, y, opt; i <= m; ++i) {scanf("%d%d%d", &opt, &x, &y); Fk[i] = {opt, x, y, 0};if(opt == 1) {assert(x >= 1); assert(x <= n);assert(y >= 0); assert(y <= 1000000000);Fk[i].old = ar[x];bit_add(x, ar[x]);bit_add(x, y);ar[x] = y;vs.eb(y);}else {assert(x >= 1); assert(x <= y); assert(y <= n);assert(opt == 2);Fk[i].old = bit_query(y) ^ bit_query(x-1);vs.eb(Fk[i].old);//printf("- %d -\n", Fk[i].old);}}sort(all(vs));vs.erase(unique(all(vs)), vs.end());ALL = vs.size();//printf("--%d\n", ALL);int cntp = 0, cnta = 0;for(int i = 1; i <= n; ++i) {cw[++cntp].p = i;cw[cntp].ip = 1;cw[cntp].x = lower_bound(all(vs), br[i]) - vs.begin() + 1;cw[cntp].id = -1;}for(int i = 1, x, y, opt; i <= m; ++i) {opt = Fk[i].opt, x = Fk[i].x, y = Fk[i].y;if(opt == 1) {cw[++cntp].ip = opt;cw[cntp].p = x;cw[cntp].id = lower_bound(all(vs), Fk[i].old) - vs.begin() + 1;cw[cntp].x = lower_bound(all(vs), y) - vs.begin() + 1;}else {cw[++cntp].ip = opt;cw[cntp].p = x-1;cw[cntp].delta = -1;cw[cntp].x = lower_bound(all(vs), Fk[i].old) - vs.begin() + 1;cw[cntp].id = ++cnta;cw[++cntp].ip = opt;cw[cntp].p = y;cw[cntp].delta = 1;cw[cntp].x = cw[cntp-1].x;cw[cntp].id = cnta;QWE[cnta] = y-x+1;}}cdq(1, cntp);for(int i = 1; i <= cnta; ++i) printf("%d\n", QWE[i] - ANS[i]);return 0;
}

CSUSTOJ 1026：强制在线树套树

/*** Copyright (C), 2019-2019, csust* FileName: try* Author:   cwolf9* Date:     19-4-16 下午6:44* Description: ${DESCRIPTION}* History:* <author>          <time>          <version>          <desc>* Cwolf9           19-4-16           2019.1            ${DESCRIPTION}*/#pragma comment(linker, "/STACK:102400000,102400000")
#include<bits/stdc++.h>
#define endl "\n"
#define fi first
#define se second
#define eb emplace_back
#define mk make_pair
#define all(x) (x).begin(), (x).end()
#define clr(a, b) memset((a),(b),sizeof((a)))
#define iis std::ios::sync_with_stdio(false);cin.tie(NULL);
using namespace std;
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int, int> pii;
typedef vector<int> VI;
typedef vector<pii> VPII;const int MXN = 1e6 + 6;
const int MXE = 2e6 + 6;
const int INF = 0x3f3f3f3f;int n, m;
struct lp {int l, r, sum;
}cw[54000007];
struct node {int opt, x, y, old;
}edge[MXE];
int ALL;
int ar[MXN], br[MXN];
int Root[MXN], NODE, yRoot[MXN];
int bit[MXN];
int lowbit(int x){return (x&(-x));}
void bit_Add(int x,int c){for(; x <= n; x += lowbit(x)) bit[x]^=c;}
int bit_Query(int x){int ANS = 0;for(; x; x -= lowbit(x)) ANS^=bit[x];return ANS;}void update(int &cur, int x, int v, int l, int r) {if(!cur) {cur = ++ NODE;cw[cur].l = cw[cur].r = cw[cur].sum = 0;}cw[cur].sum += v;if(l == r) return ;int mid = (l + r) >> 1;if(x <= mid) update(cw[cur].l, x, v, l, mid);else update(cw[cur].r, x, v, mid+1, r);
}
void update(int &cur, int old, int x, int l, int r, int p) {cur = ++ NODE;cw[cur] = cw[old];++ cw[cur].sum;if(l == r) return ;int mid = (l + r) >> 1;if(x <= mid) update(cw[cur].l, cw[old].l, x, l, mid, p);else update(cw[cur].r, cw[old].r, x, mid+1, r, p);
}
int Find(int l, int r, int aim, int rt) {if(l == r) return cw[rt].sum;int mid = (l + r) >> 1;if(aim <= mid) return Find(l, mid, aim, cw[rt].l);else return Find(mid+1, r, aim, cw[rt].r);
}
int query(int l, int r, int aim) {int ans = Find(1, ALL, aim, yRoot[r]) - Find(1, ALL, aim, yRoot[l]);for(; r; r -= lowbit(r)) ans += Find(1, ALL, aim, Root[r]);for(; l; l -= lowbit(l)) ans -= Find(1, ALL, aim, Root[l]);return ans;
}
int main() {
#ifndef ONLINE_JUDGEfreopen("/home/cwolf9/CLionProjects/newOne/in.txt", "r", stdin);//freopen("/home/cwolf9/CLionProjects/newOne/out.txt", "w", stdout);
#endifscanf("%d%d", &n, &m);VI vs; vs.eb(-1); cw[0] = {0,0,0};for(int i = 1; i <= n; ++i) scanf("%d", &ar[i]), vs.eb(ar[i]), bit_Add(i, ar[i]), br[i] = ar[i];for(int i = 1, opt, x, y; i <= m; ++i) {scanf("%d%d%d", &opt, &x, &y);edge[i] = {opt, x, y};if(opt == 1) {vs.eb(y);}}sort(all(vs));vs.erase(unique(all(vs)), vs.end());ALL = vs.size();for(int i = 1, tmp; i <= n; ++i) {tmp = lower_bound(all(vs), br[i]) - vs.begin();update(yRoot[i], yRoot[i-1], tmp, 1, ALL, 1);assert(tmp >= 1); assert(tmp <= ALL);}int aim = lower_bound(all(vs), 8) - vs.begin(), last = 0;//printf("---- %d\n", Find(1, ALL, aim, yRoot[6]) - Find(1, ALL, aim, yRoot[2]));for(int i = 1, opt, x, y; i <= m; ++i) {opt = edge[i].opt, x = edge[i].x, y = edge[i].y;if(opt == 1) {x = (x + last) % n + 1;int tmp1 = lower_bound(all(vs), ar[x]) - vs.begin();int tmp2 = lower_bound(all(vs), y) - vs.begin();assert(tmp1 >= 1); assert(tmp1 <= ALL);assert(tmp2 >= 1); assert(tmp2 <= ALL);bit_Add(x, ar[x]);bit_Add(x, y);ar[x] = y;while(x <= n) {update(Root[x], tmp1, -1, 1, ALL);update(Root[x], tmp2, 1, 1, ALL);x += lowbit(x);}}else {x = (x + last) % n + 1;y = (y + last) % n + 1;if(x > y) swap(x, y);int ans = bit_Query(y)^bit_Query(x-1);int tmp = lower_bound(all(vs), ans) - vs.begin();if(tmp == vs.size() || vs[tmp] != ans) {last = y - x + 1;}else {assert(tmp >= 1); assert(tmp <= ALL);last = (y - x + 1 - query(x - 1, y, tmp));}printf("%d\n", last);}}return 0;
}

整体二分

动态区间第k小

动态区间第k小的n种解法：here
洛谷P2617, ZOJ2112
练习题：洛谷P3332
就很板子的题。
整体二分类似于一些决策单调性的分治，可以解决诸多区间第\(k\)小的问题。
\(solve(l,r,L,R)\)表示当前值域为\([l,r]\)，当前操作为\([L,R]\)。

我们要对所有操作按照他们对应值域区间进行划分，并递归分治。分治层数只与值域区间相关，我们是带着和这个值域相关的询问向下分治。所以整体二分的复杂度也是很稳定的。

枚举\([L,R]\)：

• 如果当前操作是更新操作：若更新的值在\([l,mid]\)内，则用树状数组在当前更新操作所代表的序列下标位置\(+1\)，并将次操作归纳到下一层左区间内；反之归纳到下一层的右区间内。
• 如果当前操作是查询操作，树状数组询问区间\([cw[i].l,cw[i].r]\)内的值域情况，如果不小于\(cw[i].k\)就归入下一层左区间内；反之减去相应大小并归纳到下一层的右区间内。
• 终止条件：值域只有一个数，更新相应答案即可。

AC_Code:260ms

const int INF = 0x3f3f3f3f;
const int MXN = 2e5 + 7;
const int MXE = 1e6 + 7;
int n, m;
int bit[MXN];
void bit_add(int x,int c){for(;x <= n+1;x += lowbit(x)) bit[x]+=c;}
int bit_query(int x){int ans = 0;for(;x;x -= lowbit(x))ans+=bit[x];return ans;}
int ar[MXN];
struct lp {int l, r, v, id, ip;
}cw[MXN], cw1[MXN], cw2[MXN];
int ans[MXN];
int stk[MXN], top;
void solve(int l, int r, int L, int R) {if(l > r || L > R) return;if(l == r) {for(int i = L; i <= R; ++i) if(cw[i].ip == 2) ans[cw[i].id] = l;return;}int mid = (l + r) >> 1, cnt1 = 0, cnt2 = 0;top = 0;for(int i = L; i <= R; ++i) {if(cw[i].ip == 1) {if(cw[i].l <= mid) bit_add(cw[i].id, cw[i].v), cw1[++cnt1] = cw[i], stk[++top] = i;else cw2[++cnt2] = cw[i];}else {int k = bit_query(cw[i].r) - bit_query(cw[i].l - 1);if(k >= cw[i].v) cw1[++cnt1] = cw[i];else cw[i].v -= k, cw2[++cnt2] = cw[i];}}for(int i = 1; i <= top; ++i) bit_add(cw[stk[i]].id, -cw[stk[i]].v);for(int i = 1; i <= cnt1; ++i) cw[L + i - 1] = cw1[i];for(int i = 1; i <= cnt2; ++i) cw[L + cnt1 + i - 1] = cw2[i];solve(l, mid, L, L + cnt1 - 1);solve(mid + 1, r, L + cnt1, R);
}
int main() {
#ifndef ONLINE_JUDGEfreopen("/home/cwolf9/CLionProjects/ccc/in.txt", "r", stdin);
//    freopen("/home/cwolf9/CLionProjects/ccc/out.txt", "w", stdout);
#endifint tim = read();while(tim --) {n = read(), m = read();int tot = 0, p = 0;for(int i = 1; i <= n; ++i) ar[i] = read(), cw[++ tot] = {ar[i], 0, 1, i, 1};char s[2];int l, r, k;for(int i = 1; i <= m; ++i) {scanf("%s", s);if (s[0] == 'Q') {l = read(), r = read(), k = read();cw[++tot] = {l, r, k, ++ p, 2};}else {l = read(), r = read();cw[++tot] = {ar[l], 0, -1, l, 1};cw[++tot] = {ar[l] = r, 0, 1, l, 1};}}solve(0, INF, 1, tot);for(int i = 1; i <= p; ++i) printf("%d\n", ans[i]);}return 0;
}

P3332 [ZJOI2013]K大数查询

坑点：会爆栈导致\(mle\)，\(mle\)是因为负数二分递归原因
mle的原因是负数分治递归下去的时候死循环了，就是(l,r)没变过

• solution1：如果现在操作里没有询问操作就break
• solution2：如果递归层数大于20层就break，因为死循环只出现在负数，只有插入会出现负数，可以肯定死循环的操作里没有循环操作
• solution3：既然负数会出问题，就把负数变成整数呗

这题和上面那题唯一的区别就是单点更新变成了区间更新，用一个可以区间加区间赋零的数据结构就行了。
可以树状数组也可以线段树（线段树就用两个lazy标记，一个区间加，一个区间赋值，注意区间赋值的lazy标记优先级比较高）
代码：1, 2
AC_Code

const int MXN = 1e5 + 7;
const int MXE = 1e6 + 7;
int n, m;
//sum[i] = sigma(ar[x])+(i+1)*sigma(delta[x])-sigma(x*delta[x])
//delta[]是差分数组
struct FenwickTree {static const int MAXN = 1e5 + 6;LL delta[MAXN], deltai[MAXN];void add1(int x, int v) {while (x <= n + 10) {delta[x] += v;x += lowbit(x);}}void add2(int x, int v) {while (x <= n + 10) {deltai[x] += v;x += lowbit(x);}}LL query1(int x) {LL sum = 0;while (x > 0) {sum += delta[x];x -= lowbit(x);}return sum;}LL query2(int x) {LL sum = 0;while (x > 0) {sum += deltai[x];x -= lowbit(x);}return sum;}void update(int l, int r, int x) {add1(l, x);add1(r + 1, -x);add2(l, l * x);add2(r + 1, -x * (r + 1));}LL range(int l, int r) {LL sum1 = l * query1(l - 1) - query2(l - 1);LL sum2 = (r + 1) * query1(r) - query2(r);return sum2 - sum1;}
}bit;
struct lp {int l, r;LL v;int id, ip;lp(){}lp(int _l, int _r, LL _v, int _id, int _ip) {l = _l, r = _r, v = _v, id = _id, ip = _ip;}
}cw[MXN], cw1[MXN], cw2[MXN];
int ans[MXN];
int stk[MXN], top;
void solve(int l, int r, int L, int R, int d) {if(d > 18) return;if(l > r || L > R) return;if(l == r) {for(int i = L; i <= R; ++i) if(cw[i].ip == 2) ans[cw[i].id] = l;return;}int mid = (l + r) / 2, cnt1 = 0, cnt2 = 0;top = 0;bool fl = false, fr = false;for(int i = L; i <= R; ++i) {if(cw[i].ip == 1) {if(cw[i].v > mid) bit.update(cw[i].l, cw[i].r, 1), cw1[++cnt1] = cw[i], stk[++top] = i;else cw2[++cnt2] = cw[i];}else {LL k = bit.range(cw[i].l, cw[i].r);if(k >= cw[i].v) cw1[++cnt1] = cw[i], fr = true;else cw[i].v -= k, cw2[++cnt2] = cw[i], fl = true;}}for(int i = 1; i <= top; ++i) bit.update(cw[stk[i]].l, cw[stk[i]].r, -1);for(int i = 1; i <= cnt2; ++i) cw[L + i - 1] = cw2[i];for(int i = 1; i <= cnt1; ++i) cw[L + cnt2 + i - 1] = cw1[i];if(fl) solve(l, mid, L, L + cnt2 - 1, d + 1);if(fr) solve(mid + 1, r, L + cnt2, R, d + 1);
}
int main() {
#ifndef ONLINE_JUDGEfreopen("/home/cwolf9/CLionProjects/ccc/in.txt", "r", stdin);
//    freopen("/home/cwolf9/CLionProjects/ccc/out.txt", "w", stdout);
#endif
//    int tim = 1;n = read(), m = read();int p = 0;int l, r, opt;LL k;for(int i = 1; i <= m; ++i) {opt = read();if (opt == 1) {l = read(), r = read(), k = read();cw[i] = lp(l, r, k, l, 1);}else {l = read(), r = read(), k = read();cw[i] = lp(l, r, k, ++ p, 2);}}solve(-n, n, 1, m, 1);for(int i = 1; i <= p; ++i) printf("%d\n", ans[i]);return 0;
}