题链：

http://www.lydsy.com/JudgeOnline/problem.php?id=1934

题解：

题目有点迷。

S向为1的点连边，为0的点向T连边，
在有关系的两个点之间连双向边。
以上边的容量都为1。
然后求最小割，最小割==最大流，dinic实现。
代码：

#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#define MAXN 350
#define INF 0x3f3f3f3f
using namespace std;
struct Edge{int to[MAXN*MAXN*2],cap[MAXN*MAXN*2],nxt[MAXN*MAXN*2],head[MAXN],ent;void Init(){ent=2; memset(head,0,sizeof(head));}void Adde(int u,int v,int w){to[ent]=v; cap[ent]=w; nxt[ent]=head[u]; head[u]=ent++;to[ent]=u; cap[ent]=0; nxt[ent]=head[v]; head[v]=ent++;}int Next(bool type,int i){return type?head[i]:nxt[i];}
}E;
int N,M,T,S;
int d[MAXN],cur[MAXN];
bool bfs(){memset(d,0,sizeof(d));queue<int>q; q.push(S); d[S]=1;while(!q.empty()){int u=q.front(); q.pop();for(int i=E.Next(1,u);i;i=E.Next(0,i)){int v=E.to[i];if(d[v]||!E.cap[i]) continue;d[v]=d[u]+1; q.push(v);}}return d[T];
}
int dfs(int u,int reflow){if(!reflow||u==T) return reflow;int flowout=0,f;for(int &i=cur[u];i;i=E.Next(0,i)){int v=E.to[i];if(d[v]!=d[u]+1) continue;f=dfs(v,min(reflow,E.cap[i]));reflow-=f; E.cap[i]-=f;flowout+=f; E.cap[i^1]+=f;if(!reflow) break;}if(!flowout) d[u]=0;return flowout;
}
int Dinic(){int flow=0;while(bfs()){memcpy(cur,E.head,sizeof(E.head));flow+=dfs(S,INF);}return flow;
}
int main()
{scanf("%d%d",&N,&M); S=N+1,T=N+2;E.Init();for(int i=1,x;i<=N;i++){scanf("%d",&x);if(x==1) E.Adde(S,i,1);else E.Adde(i,T,1);}for(int i=1,a,b;i<=M;i++){scanf("%d%d",&a,&b);E.Adde(a,b,1);E.Adde(b,a,1);}int ans=Dinic();printf("%d",ans);return 0;
}