luogu P5785 [SDOI2012]任务安排 斜率dp+二分

Posted 2021-02-19 qingyuyyyyy

//k可能是负数，所以斜率不在有单调性
//新加的点的横坐标还是单调递增的，但可能不是严格的
//在查询的时候：只能二分查找
//在插入的时候：将队尾不在凸包上的点删掉

//f[j]=(sumt[i]+s)*sumc[j]+f[i]-sumt[i]*sumc[i]-s*sumc[n]
//f[i]=f[j]-(sumt[i]+s)sumc[j]+sumt[i]*sumc[i]+s*sumc[n]
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
typedef long long LL;
const int N = 3e5 + 10;
int n, s;
LL sc[N], st[N];
LL f[N];
int q[N];
double Y(int i)
{
	return (double)f[i];
}
double X(int i)
{
	return (double)sc[i];
}
double slope(int i,int j)
{
	if(X(i)-X(j)<-1e5)
		return 1e9;
	return (double)(Y(i)-Y(j))/(X(i)-X(j));
}
int main()
{
	cin>>n>>s;
	for(int i=1; i<=n; i++)
	{
		cin>>st[i]>>sc[i];
		st[i]+=st[i-1];
		sc[i]+=sc[i-1];
	}
	int hh=0,tt=0;
	//f[j]=(sumt[i]+s)*sumc[j]+f[i]-sumt[i]*sumc[i]-s*sumc[n]
	//f[i]=f[j]-(sumt[i]+s)sumc[j]+sumt[i]*sumc[i]+s*sumc[n]
	for(int i=1; i<=n; i++)
	{
		int l=hh,r=tt;
		while(l<r)
		{
			int mid=l+r>>1;
			if(slope(q[mid+1],q[mid])>st[i]+s)
				r=mid;
			else
				l=mid+1;
		}
		f[i]=f[q[l]]-(st[i]+s)*sc[q[l]]+st[i]*sc[i]+s*sc[n];
		while(hh<tt&&slope(i,q[tt-1])<=slope(q[tt],q[tt-1]))
			tt--;
		q[++tt]=i;
	}
	cout<<f[n]<<endl;
}