poj 3762 The Bonus Salary! 需离散化(二)
void end(int s,int t) { int u, sum = inf; for(u=t; u!=s; u=pre[u]) { sum = min(sum,Edge[path[u]].cap); } max_flow+=sum; //记录最大流 for(u = t; u != s; u=pre[u]) { Edge[path[u]].cap -= sum; Edge[path[u]^1].cap += sum; ans += sum*Edge[path[u]].cost; // cost记录的为单位流量费用，必须得乘以流量。 } } int main() { int n,k,s,t; scanf("%d%d",&n,&k); memset(head,-1,sizeof(head)); memset(z,0,sizeof(z)); NE=ans=max_flow=s=0; int num=0,h,m,ss; for(int i=0; i