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;

}