【洛谷 P3299】 [SDOI2013]保护出题人 （凸包，三分，斜率优化）

题目链接
易得第\(i\)关的最小攻击力为\(\max_{j=1}^i\frac{sum[i]-sum[j-1]}{x+d*(i-j)}\)
十分像一个斜率式，于是看作一个点\(P(x+d*i,sum[i])\)和点\(Q(d*j,sum[j-1])\)的斜率
于是就是求当前\(i\)的点\(P\)和之前的所有点\(Q\)的最大斜率，显然有最大斜率的点在凸包上而且有单峰。
于是用单调队列维护凸包，在凸包上三分斜率最大值。

#include <cstdio>
#include <algorithm>
#define ll long long
using namespace std;
const int MAXN = 100010;
inline ll read(){
    ll s = 0;
    int w = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9'){ if(ch == '-') w = -1; ch = getchar(); }
    while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); }
    return s * w;
}
int n, top;
ll d;
ll a[MAXN], x[MAXN], sum[MAXN];
int st[MAXN];
double k(int i, int j){
    return ((double)sum[i - 1] - sum[j - 1]) / (i * d - j * d);
}
double calc(int i, int j){
    return ((double)sum[i] - sum[j - 1]) / (x[i] + d * (i - j));
}
double ans;
int main(){
    n = read(); d = read();
    for(int i = 1; i <= n; ++i){
       a[i] = read(); x[i] = read();
       sum[i] = sum[i - 1] + a[i];
    }
    for(int i = 1; i <= n; ++i){
       while(top > 1 && k(st[top - 1], st[top]) > k(st[top - 1], i)) --top;
       st[++top] = i;
       int l = 1, r = top, mid1, mid2, p;
       while(l != r){
          p = (r - l + 1) / 3;
          mid1 = l + p - 1;
          mid2 = l + p + p - 1;
          if(mid1 == mid2) break;
          if(calc(i, st[mid1]) < calc(i, st[mid2])) l = mid1 + 1;
          else r = mid2 - 1;
       }
       ans += max(calc(i, st[l]), calc(i, st[r]));
    }
    printf("%.0lf\n", ans);
    return 0;
}