ARC065

C - 白昼夢 / Daydream

直接递推就好

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {res = 0;T f = 1;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {res = res * 10 +c - '0';c = getchar();}res *= f;
}
template<class T>
void out(T x) {if(x < 0) {x = -x;putchar('-');}if(x >= 10) {out(x / 10);}putchar('0' + x % 10);
}
string s;
int dp[MAXN];
void Solve() {cin >> s;dp[0] = 1;for(int i = 1 ; i <= s.length() ; ++i) {if(i >= 5) {dp[i] |= (dp[i - 5] && s.substr(i - 5,5) == "dream");dp[i] |= (dp[i - 5] && s.substr(i - 5,5) == "erase");}if(i >= 6) {dp[i] |= (dp[i - 6] && s.substr(i - 6,6) == "eraser");}if(i >= 7) {dp[i] |= (dp[i - 7] && s.substr(i - 7,7) == "dreamer");}}if(dp[s.length()]) puts("YES");else puts("NO");
}
int main() {
#ifdef ivorysifreopen("f1.in","r",stdin);
#endifSolve();return 0;
}

D - 連結 / Connectivity

把图两次染色，然后标号，会得到一个点对，这个点对相同的点才在两个图里都连通

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {res = 0;T f = 1;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {res = res * 10 +c - '0';c = getchar();}res *= f;
}
template<class T>
void out(T x) {if(x < 0) {x = -x;putchar('-');}if(x >= 10) {out(x / 10);}putchar('0' + x % 10);
}
map<pii,int> zz;
int N,K,L;
int fa[MAXN],col[MAXN][2],tot;
int getfa(int x) {return fa[x] == x ? x : fa[x] = getfa(fa[x]);
}
void Solve() {read(N);read(K);read(L);for(int i = 1 ; i <= N ; ++i) fa[i] = i;int a,b;for(int i = 1 ; i <= K ; ++i) {read(a);read(b);fa[getfa(a)] = getfa(b);}tot = 0;for(int i = 1 ; i <= N ; ++i) {if(getfa(i) == i) col[i][0] = ++tot;}for(int i = 1 ; i <= N ; ++i) col[i][0] = col[getfa(i)][0];for(int i = 1 ; i <= N ; ++i) fa[i] = i;for(int i = 1 ; i <= L ; ++i) {read(a);read(b);fa[getfa(a)] = getfa(b);}tot = 0;for(int i = 1 ; i <= N ; ++i) {if(getfa(i) == i) col[i][1] = ++tot;}for(int i = 1 ; i <= N ; ++i) col[i][1] = col[getfa(i)][1];for(int i = 1 ; i <= N ; ++i) {zz[mp(col[i][0],col[i][1])]++;}for(int i = 1 ; i <= N ; ++i) {out(zz[mp(col[i][0],col[i][1])]);space;}enter;
}
int main() {
#ifdef ivorysifreopen("f1.in","r",stdin);
#endifSolve();return 0;
}

E - へんなコンパス / Manhattan Compass

曼哈顿转切比雪夫

然后把点对排序，每个点用横坐标相连就是

\(x_{a} - x_{b} = D\)

并且$ -D<= y_{a} - y_{b} <= D$

这是一段区间，我们可以把区间里的点两两相连，再和a点相连，是等价的

然后记录每个点连出去的边有多少，和初始的两个点连通的所有点之间的边数加起来就是答案

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {res = 0;T f = 1;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {res = res * 10 +c - '0';c = getchar();}res *= f;
}
template<class T>
void out(T x) {if(x < 0) {x = -x;putchar('-');}if(x >= 10) {out(x / 10);}putchar('0' + x % 10);
}
struct node {int to,next;
}E[MAXN * 10];
int head[MAXN],sumE;
int N,a,b;
int x[MAXN],y[MAXN],D,id[MAXN],cnt[MAXN];
pii p[MAXN],tmp[MAXN];
int st[MAXN],ed[MAXN];
int fa[MAXN];
int getfa(int x) {return fa[x] == x ? x : fa[x] = getfa(fa[x]);
}
void Merge(int x,int y) {fa[getfa(x)] = getfa(y);
}
void AddE(int on) {for(int i = 1 ; i <= N ; ++i) {id[i] = i;}sort(id + 1,id + N + 1,[](int a,int b){return p[a] < p[b];});memset(st,0,sizeof(st));memset(ed,0,sizeof(ed));for(int i = 1 ; i <= N ; ++i) {tmp[i] = p[id[i]];}for(int i = 1 ; i <= N ; ++i) {int l = lower_bound(tmp + 1,tmp + N + 1,mp(tmp[i].fi + D,tmp[i].se - D + on)) - tmp;int r = upper_bound(tmp + 1,tmp + N + 1,mp(tmp[i].fi + D,tmp[i].se + D - on)) - tmp - 1;if(l <= r) {st[l]++;ed[r]++;Merge(id[i],id[l]);cnt[id[i]] += r - l + 1;}}int pre = 0;for(int i = 1 ; i < N ; ++i) {pre += st[i];pre -= ed[i];if(pre) Merge(id[i],id[i + 1]);}
}
void Solve() {read(N);read(a);read(b);for(int i = 1 ; i <= N ; ++i) {read(x[i]);read(y[i]);fa[i] = i;}D = abs(x[a] - x[b]) + abs(y[a] - y[b]);for(int i = 1 ; i <= N ; ++i) p[i] = mp(x[i] + y[i],x[i] - y[i]);AddE(0);for(int i = 1 ; i <= N ; ++i) swap(p[i].fi,p[i].se);AddE(1);int64 ans = 0;for(int i = 1 ; i <= N ; ++i) {if(getfa(i) == getfa(a)) ans += cnt[i];}out(ans);enter;
}
int main() {
#ifdef ivorysifreopen("f1.in","r",stdin);
#endifSolve();return 0;
}

F - シャッフル / Shuffling

就是对于一个区间，记录能用a个0和b个1

然后对于某个位置填0还是填1

由于能用的数是总和是确定的，我们只要记录当前有a个0没用就可以了

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 3005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {res = 0;T f = 1;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {res = res * 10 +c - '0';c = getchar();}res *= f;
}
template<class T>
void out(T x) {if(x < 0) {x = -x;putchar('-');}if(x >= 10) {out(x / 10);}putchar('0' + x % 10);
}
const int MOD = 1000000007;
int ri[MAXN],f[2][MAXN];
int N,M;
char s[MAXN];
int inc(int a,int b) {return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {return 1LL * a * b % MOD;
}
void update(int &x,int y) {x = inc(x,y);
}
void Solve() {read(N);read(M);scanf("%s",s + 1);int l,r;for(int i = 1 ; i <= M ; ++i) {read(l);read(r);ri[l] = max(ri[l],r);}int p = 0,all = 0;int cur = 0;f[cur][0] = 1;for(int i = 1 ; i <= N ; ++i) {if(p < i - 1) {p = i - 1;}if(ri[i] > p) {all += ri[i] - p;int a = 0,b = 0;for(int j = p + 1 ; j <= ri[i] ; ++j) {if(s[j] == '0') ++a;else ++b;}p = ri[i];memset(f[cur ^ 1],0,sizeof(f[cur ^ 1]));for(int j = a ; j <= N ; ++j) f[cur ^ 1][j] = f[cur][j - a];cur ^= 1;}if(!all) continue;memset(f[cur ^ 1],0,sizeof(f[cur ^ 1]));for(int j = 0 ; j <= all ; ++j) {if(j < all) update(f[cur ^ 1][j],f[cur][j]);if(j >= 1) update(f[cur ^ 1][j - 1],f[cur][j]);}--all;cur ^= 1;}out(f[cur][0]);enter;
}
int main() {
#ifdef ivorysifreopen("f1.in","r",stdin);
#endifSolve();return 0;
}