题目：

一个斜率优化+CDQ好题

BZOJ2149

分析：

先吐槽一下题意：保留房子反而要给赔偿金是什么鬼哦……

第一问是一个经典问题。直接求原序列的最长上升子序列是错误的。比如\(\{1,2,2,3\}\)，选择\(\{1,2,3\}\)不改变后会发现无论如何修改都无法变成一个严格上升序列。只能选择\(\{1,2\}\)，把原序列改成\(\{1,2,3,4\}\)。

考虑对于两个数\(a_i\)和\(a_j(j<i)\)，\(a_i\)能接在\(a_j\)后面的充要条件是\(a_i-a_j\geq i-j\)（这样中间才能塞下\(i-j-1\)个数形成上升序列）。移项得到\(a_i-i\geq a_j-j\)，所以应该把每个数减去它的编号作为权值然后求最长非降子序列。由于要求美观度为正整数，所以若\(a_i-i<0\)，则\(i\)不能作为序列的开端。下面的代码展示了\(O(nlog_2n)\)求法（其中\(c[i]=a[i]-i\)，\(f[i]\)表示以\(i\)结尾的最长非降子序列的长度）。

int solve()
{int ans = 0;memset(tmp, INF, sizeof(int[n + 1]));for (int i = 1; i <= n; i++){if (c[i] < 0)f[i] = 0;else{int pos = upper_bound(tmp + 1, tmp + ans + 1, c[i]) - tmp;tmp[pos] = c[i];ans = max(ans, pos);f[i] = pos;}v[f[i]].push_back(i);}return ans;
}

然后来看第二问。设\(dp[i]\)为将前\(i\)个数变成单调上升序列的最小总花费。则\(dp[i]\)可以由\(dp[j]\)转移而来的必要条件是\(i>j\)，\(a[i]-i>a[j]-j\)且\(f[i]=f[j]+1\)（若\(f[i]>f[j]+1\)，则不满足“保留最多的旧房子”；若\(f[i]<f[j]+1\)，说明你\(f[i]\)算错了）。

转移时，最优解显然是把\(a[k](j<k<i)\)变成一个以\(a[j]+1\)为首项，公差为\(1\)的等差数列。由于\(a[i]-i>a[j]-j\)，所以改完以后一定有\(a[i-1]<a[i]\)

\[dp[i]=min\{dp[j]+\frac{[a[j]+1+a[j]+(i-j-1)]\times(i-j-1)}{2}+a[i]+b[i]\}\]

整理一下，得到：

\[dp[i]=min\{dp[j]+a[j]\times(i-j-1)+\frac{i(i-1)}{2}+\frac{j(j+1)}{2}+-ij+a[i]+b[i]\}\]

可以根据\(f[i]\)分层，一起处理所有\(f[j]=k-1\)的\(j\)对\(f[i]=k\)的\(i\)的贡献。下面考虑每一层的情况。

未完待续……

代码：

方便起见，在序列首加一个\(0\)（\(a[0]=f[0]=0\)）。这样可以保证改造后美观度为正（因为\(f[i]=1\)的\(dp[i]\)必然从\(dp[0]\)转移而来）；在序列尾加一个无穷大作为\(a[n+1]\)，\(dp[n+1]-a[n+1]\)即为答案。

#include <cstdio>
#include <algorithm>
#include <cctype>
#include <cstring>
#include <vector>
using namespace std;namespace zyt
{template<typename T>inline void read(T &x){char c;bool f = false;x = 0;doc = getchar();while (c != '-' && !isdigit(c));if (c == '-')f = true, c = getchar();dox = x * 10 + c - '0', c = getchar();while (isdigit(c));if (f)x = -x;}template<typename T>inline void write(T x){static char buf[20];char *pos = buf;if (x < 0)putchar('-'), x = -x;do*pos++ = x % 10 + '0';while (x /= 10);while (pos > buf)putchar(*--pos);}typedef long long ll;typedef long double ld;const int N = 1e5 + 10, INF = 0x3f3f3f3f;const ll LINF = 0x3f3f3f3f3f3f3f3fLL;int n, a[N], b[N], c[N], f[N], tmp[N];ll dp[N];vector<int> v[N];int solve(){int ans = 0;memset(tmp, INF, sizeof(int[n + 1]));for (int i = 1; i <= n; i++){if (c[i] < 0)f[i] = 0;else{int pos = upper_bound(tmp + 1, tmp + ans + 1, c[i]) - tmp;tmp[pos] = c[i];ans = max(ans, pos);f[i] = pos;}v[f[i]].push_back(i);}v[0].push_back(0);return ans;}inline ll x(const int i){return i - a[i];}inline ll y(const int i){return dp[i] - (ll)(i + 1) * a[i] + (ll)i * (i + 1) / 2;}inline ld ratio(const int i, const int j){if (x(i) == x(j))return y(i) < y(j) ? -LINF : LINF;elsereturn (ld)(y(i) - y(j)) / (x(i) - x(j));}struct node{int pos;bool type;bool operator < (const node &b) const{return pos < b.pos;}}arr[N];const int CHANGE = 0, QUERY = 1;void CDQ(const int l, const int r){if (l == r)return;int mid = (l + r) >> 1, i = l, j = mid + 1, k = l;static node tmp[N];static int st[N];CDQ(l, mid), CDQ(mid + 1, r);int top = 0;while (i <= mid && j <= r){if (x(arr[i].pos) >= x(arr[j].pos)){if (arr[i].type == CHANGE){while (top > 1 && ratio(st[top - 2], st[top - 1]) < ratio(st[top - 1], arr[i].pos))--top;st[top++] = arr[i].pos;}tmp[k++] = arr[i++];}else{if (arr[j].type == QUERY && top){int l = 0, r = top - 2, ans = top - 1;while (l <= r){int mid = (l + r) >> 1;if (ratio(st[mid], st[mid + 1]) < arr[j].pos)r = mid - 1, ans = mid;elsel = mid + 1;}dp[arr[j].pos] = min(dp[arr[j].pos], dp[st[ans]] + (ll)((a[st[ans]] << 1) + arr[j].pos - st[ans]) * (arr[j].pos - st[ans] - 1) / 2+ a[arr[j].pos] + b[arr[j].pos]);}tmp[k++] = arr[j++];}}while (i <= mid)tmp[k++] = arr[i++];while (j <= r){if (arr[j].type == QUERY && top){int l = 0, r = top - 2, ans = top - 1;while (l <= r){int mid = (l + r) >> 1;if (ratio(st[mid], st[mid + 1]) < arr[j].pos)r = mid - 1, ans = mid;elsel = mid + 1;}dp[arr[j].pos] = min(dp[arr[j].pos], dp[st[ans]] + (ll)((a[st[ans]] << 1) + arr[j].pos - st[ans]) * (arr[j].pos - st[ans] - 1) / 2+ a[arr[j].pos] + b[arr[j].pos]);}tmp[k++] = arr[j++];}memcpy(arr + l, tmp + l, sizeof(node[r - l + 1]));}int work(){read(n);for (int i = 1; i <= n; i++)read(a[i]), c[i] = a[i] - i;for (int i = 1; i <= n; i++)read(b[i]);a[++n] = INF;c[n] = INF;int ans = solve();write(ans - 1), putchar(' ');memset(dp, INF, sizeof(ll[n + 1]));dp[0] = 0;for (int i = 1; i <= ans; i++){int cnt = 0;for (int j = 0; j < v[i - 1].size(); j++)if (dp[v[i - 1][j]] < LINF)arr[++cnt] = (node){v[i - 1][j], CHANGE};for (int j = 0; j < v[i].size(); j++)arr[++cnt] = (node){v[i][j], QUERY};sort(arr + 1, arr + cnt + 1);CDQ(1, cnt);}write(dp[n] - a[n] - b[n]);return 0;}
}
int main()
{return zyt::work();
}