poj 3308 网络流

由行向列建图，增加两个源汇点，源点向行连边，容量为该行的花费，列向汇点连边，容量为该列的代价，一个伞兵的位置为x，y，

则从x行向y列连一天容量为无穷大的边，最后求最大流（最小割的容量）即可

View Code

#include<cstdio>
#include<cstring>
#include<cmath>
const int MAX=100005;
const double INF=1000000000;
struct
{
    int v,next;
    double c;
}edge[1000000];
int E,head[MAX];
int gap[MAX],cur[MAX];
int pre[MAX],dis[MAX];
void add_edge(int s,int t,double c,double cc)
{
    /*加边的时候同时加两条，
    一条正的，一条反的，
    一般情况下反的容量是0 */
    edge[E].v=t; edge[E].c=c;
    edge[E].next=head[s];
    head[s]=E++;
    edge[E].v=s; edge[E].c=cc;
    edge[E].next=head[t];
    head[t]=E++;
}
double min(double a,double b){return (a==-1||b<a)?b:a;}
double SAP(int s,int t,int n)
{
    memset(gap,0,sizeof(gap));
    memset(dis,0,sizeof(dis));
    int i;
    for(i=0;i<n;i++)cur[i]=head[i];
    int u=pre[s]=s,v;
    double maxflow=0,aug=-1;
    gap[0]=n;    
    while(dis[s]<n)
    {
loop:    for(i=cur[u];i!=-1;i=edge[i].next)
        {
            v=edge[i].v;
            if(edge[i].c>0&&dis[u]==dis[v]+1)
            {
                aug=min(aug,edge[i].c);
                pre[v]=u;
                cur[u]=i;
                u=v;
                if(u==t)
                {
                    for(u=pre[u];v!=s;v=u,u=pre[u])
                    {
                        edge[cur[u]].c-=aug;
                        edge[cur[u]^1].c+=aug;
                    }
                    maxflow+=aug;
                    aug=-1;
                }
                goto loop;
            }
        }
        int mindis=n;
        for(i=head[u];i!=-1;i=edge[i].next)
        {
            v=edge[i].v;
            if(edge[i].c>0&&dis[v]<mindis)
            {
                cur[u]=i;
                mindis=dis[v];
            }
        }
        if((--gap[dis[u]])==0)break;
        gap[dis[u]=mindis+1]++;
        u=pre[u];
    }
    return maxflow;
}
int main()
{

    int t,m,n,L;
    scanf("%d",&t);
    while(t--)
    {    
        memset(head,-1,sizeof(head));
        E=0;
        scanf("%d%d%d",&m,&n,&L);
        int i,j;double co;
        for(i=1;i<=m;i++)
        {
            scanf("%lf",&co);
            add_edge(0,i,log(co),0);
        }
        for(i=1;i<=n;i++)
        {
            scanf("%lf",&co);
            add_edge(m+i,m+n+1,log(co),0);
        }
        for(i=1;i<=L;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            add_edge(x,y+m,INF,0);
        }
        printf("%.4lf\n",exp(SAP(0,n+m+1,n+m+2)));
    }
    return 0;
}