BZLJ 3203 Luogu P3299 [SDOI2013]保护出题人 (凸包、斜率优化、二分)

惊了，我怎么这么菜啊。。

题目链接: (bzoj)https://www.lydsy.com/JudgeOnline/problem.php?id=3203

(luogu)https://www.luogu.org/problemnew/show/P3299

题解: 先讲正常做法。

设\(S_i\)为\(i\)的前缀和，则显然第\(i\)次答案为\(\max^i_{j=1} \frac{S_i-S_{j-1}}{x_i+id-jd}\)

那么很显然就是要求从一个点\((x_i+id,S_i)\)到\((jd,S_{j-1})\)的斜率最大值啊。。三分凸壳就行了啊。。想什么呢。。。

考虑斜率优化(我就是陷入套路无法自拔的垃圾)，首先为了方便我们把输入的\(x_i\)加上\(id\)并记作\(x_0\), 目前的总和记作\(S\), \(1\)到\((i-1)\)的和记作\(S_i\)(和上面定义不同)，考虑\(i\)不比\(j\)差(\(i<j\))的条件: \[\frac{S-S_i}{x_0-id}>\frac{S-S_j}{x_0-jd}\]展开后解得\[x_0\ge\frac{iS-jS-iS_j+jS_i}{S_j-S_i}d\]考虑三个点\(i<j<k\), \(i\)不比\(j\)劣的条件是\[x_0\ge\frac{iS-jS-iS_j+jS_i}{S_j-S_i}d\] \(k\)不比\(j\)劣的条件是\[x_0\le\frac{jS-kS-jS_k+kS_j}{S_k-S_j}d\]若后者大于等于前者则无论\(x_0\)为何值此两个条件至少满足一个，\(j\)无用。

然后尝试化这个式子: \[\frac{jS-kS-jS_k+kS_j}{S_k-S_j}\ge \frac{iS-jS-iS_j+jS_i}{S_j-S_i}\] \[iS_jS_k-iS_j^2-jS_iS_k+jS_iS_j+jSS_k-jSS_j-iSS_k+iSS_j\le jS_kS_j-jS_iS_k-kS_j^2+kS_iS_j+kSS_j-kSS_i-jSS_j+jSS_i\] \[(i-j)S_jS_k+(k-i)S_j^2+(j-k)S_iS_j+(j-i)SS_k+(i-k)SS_j+(k-j)SS_i\le 0\] 尝试因式分解 \[(S_j-S)(iS_k-jS_k+kS_j-iS_j+jS_i-kS_i)\le 0\] 因为\(S_j-S\)显然小于\(0\), 所以\[iS_k-jS_k+kS_j-iS_j+jS_i-kS_i\ge 0\] 这个东西一看就是可以拆添项的: \[iS_k-jS_k+kS_j-jS_j-iS_j+jS_j+iS_k-jS_k\le 0\] \[(j-k)(S_i-S_j)-(i-j)(S_j-S_k)\le 0\] \[\frac{S_i-S_j}{i-j}\le \frac{S_j-S_k}{j-k}\]

这是\(j\)无用的条件，所以只需要维护斜率不增的上凸壳即可

……

我真的蠢到一定境界了

代码

当然是我的垃圾做法的代码

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cassert>
#define llong long long
using namespace std;

struct Point
{
    double x,y;
    Point() {}
    Point(llong _x,llong _y) {x = _x,y = _y;}
};
const int N = 1e5;
Point ch[N+3];
double s[N+3];
double a[N+3];
double qr[N+3];
int n,tp;
double d;

void insertpoint(Point x)
{
    while(tp>1 && (ch[tp].y-ch[tp-1].y)*(x.x-ch[tp].x)>=(x.y-ch[tp].y)*(ch[tp].x-ch[tp-1].x)) {tp--;}
    tp++; ch[tp] = x;
//  printf("CH: size=%d ",tp);
//  for(int i=1; i<=tp; i++) printf("(%lf %lf) ",ch[i].x,ch[i].y); puts("");
}

double query(double x,double sum)
{
//  printf("query(%lf %lf)\n",x,sum);
    int left = 1,right = tp;
    while(left<right)
    {
        int mid = (left+right+1)>>1;
        bool ok = x*(ch[mid].y-ch[mid-1].y)<=(ch[mid-1].x*ch[mid].y-ch[mid].x*ch[mid-1].y+ch[mid].x*sum-ch[mid-1].x*sum)*d ? true : false;
        if(ok) {left = mid;}
        else {right = mid-1;}
    }
//  printf("left=%d\n",left);
    double ret = (sum-ch[left].y)/(x-ch[left].x*d);
    return ret;
}

int main()
{
    scanf("%d%lf",&n,&d);
    for(int i=1; i<=n; i++)
    {
        scanf("%lf%lf",&a[i],&qr[i]);
        s[i] = s[i-1]+a[i-1];
        qr[i] += i*d;
    }
    s[n+1] = s[n]+a[n];
    double sans = 0.0;
    for(int i=1; i<=n; i++)
    {
        insertpoint(Point(i,s[i]));
        double ans = query(qr[i],s[i+1]);
        sans += ans;
//      printf("i%d ans%lf\n",i,ans);
    }
    printf("%lld\n",(llong)(sans+0.5));
    return 0;
}

原文地址：https://www.cnblogs.com/suncongbo/p/11116288.html