hdu 3061 (最大权闭合图)

分析：城池之间有依赖关系，汇点与能获得兵力的城池连接，容量为可以获得的兵力，损耗兵力的城池与汇点连接容量为损耗的兵力，有依赖关系的城池间连边，容量为无穷大，跑网络流求出的最小割就是损耗的最小兵力，，，

#include<stdio.h>
#include<string.h>
const int N=510;
const int inf=0x3fffffff;
int dis[N],gap[N],start,end,ans,head[N],num;
struct edge
{
	int st,ed,flow,next;
}e[N*N];
void addedge(int x,int y,int w)
{
	e[num].st=x;e[num].ed=y;e[num].flow=w;e[num].next=head[x];head[x]=num++;
	e[num].st=y;e[num].ed=x;e[num].flow=0;e[num].next=head[y];head[y]=num++;
}
int dfs(int u,int minflow)
{
    if(u==end)return minflow;
    int i,v,f,min_dis=ans-1,flow=0;
    for(i=head[u];i!=-1;i=e[i].next)
    {
        v=e[i].ed;
        if(e[i].flow<=0)continue;
        if(dis[v]+1==dis[u])
        {
            f=dfs(v,e[i].flow>minflow-flow?minflow-flow:e[i].flow);
            e[i].flow-=f;
            e[i^1].flow+=f;
            flow+=f;
            if(flow==minflow)break;
            if(dis[start]>=ans)return flow;
        }
        min_dis=min_dis>dis[v]?dis[v]:min_dis;
    }
    if(flow==0)
    {
        if(--gap[dis[u]]==0)
            dis[start]=ans;
        dis[u]=min_dis+1;
        gap[dis[u]]++;
    }
    return flow;
}
int isap()
{
    int maxflow=0;
    memset(gap,0,sizeof(gap));
    memset(dis,0,sizeof(dis));
    gap[0]=ans;
    while(dis[start]<ans)
        maxflow+=dfs(start,inf);
	//printf("%d\n",maxflow);
    return maxflow;
}
int main()
{
	int i,x,y,n,m,k;
	while(scanf("%d%d",&n,&m)!=-1)
	{
		memset(head,-1,sizeof(head));
		num=0;start=0;end=n+1;ans=end+1;k=0;
		for(i=1;i<=n;i++)
		{
			scanf("%d",&x);
			if(x>=0)
			{
			   addedge(start,i,x);
			   k+=x;
			}
			else addedge(i,end,-x);
		}
		for(i=1;i<=m;i++)
		{
			scanf("%d%d",&x,&y);
			addedge(x,y,inf);
		}
		printf("%d\n",k-isap());
	}
	return 0;
}