BZOJ 2726: [SDOI2012]任务安排

斜率优化，交叉相乘

#include<cstdio>
#include<algorithm>
using namespace std;
int t[1000005],f[1000005];
long long F[1000005],G[1000005],Sum[1000005],C[1000005];
struct node{
	long long k,b;
}stack[1000005];
double check(node a,node b){
	return ((double)a.b-b.b)/(b.k-a.k);
}
int main(){
	int n,S;
	scanf("%d%d",&n,&S);
	for (int i=1; i<=n; i++){
		scanf("%d%d",&t[i],&f[i]);
		Sum[i]=Sum[i-1]+t[i];
		C[i]=C[i-1]+f[i];
	}
	int top=0;
	for (int i=0; i<=n; i++){
		int L=1,R=top;
		while (L<R){
			int mid=(L+R)>>1;
			double l,r;
			if (mid==1) l=-1ll<<60;
			else l=check(stack[mid-1],stack[mid]);
			if (mid==top) r=1ll<<60;
			else r=check(stack[mid+1],stack[mid]);
			if (Sum[i]>=l && Sum[i]<=r){
				L=mid;
				break;
			}
			else if (Sum[i]<l) R=mid-1;
			else L=mid+1;
		}
		F[i]=stack[L].k*Sum[i]+stack[L].b;
		F[i]+=Sum[i]*C[n];
		G[i]=F[i]+(-Sum[i]+S)*(C[n]-C[i]);
		node line=(node){-C[i],G[i]};
		while (top>=2 &&(stack[top].b-stack[top-1].b)*(line.k-stack[top].k)<=
						(line.b-stack[top].b)*(stack[top].k-stack[top-1].k))top--;
		stack[++top]=line;
	}
	printf("%lld\n",F[n]);
	return 0;
}