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  • 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
    ",F[n]);
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/silenty/p/9797835.html
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