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
",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
",k-isap());
}
return 0;
}