考虑源点为同意，汇点为反对，那么只要源点向同意的连边，不同意的向汇点连边，求个最小割就是答案
然后考虑朋友之间怎么办，我们令朋友之间连双向边。这样不管怎么割都能对应一种选择情况。那么还是求一个最小割就行了
1 //minamoto 2 #include<iostream> 3 #include<cstdio> 4 #include<cstring> 5 #include<queue> 6 #define inf 0x3f3f3f3f 7 using namespace std; 8 #define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++) 9 char buf[1<<21],*p1=buf,*p2=buf; 10 inline int read(){ 11 #define num ch-'0' 12 char ch;bool flag=0;int res; 13 while(!isdigit(ch=getc())) 14 (ch=='-')&&(flag=true); 15 for(res=num;isdigit(ch=getc());res=res*10+num); 16 (flag)&&(res=-res); 17 #undef num 18 return res; 19 } 20 const int N=5005,M=500005; 21 int head[N],Next[M],ver[M],edge[M],tot=1; 22 int S,T,dep[N],cur[N],n,m; 23 queue<int> q; 24 inline void add(int u,int v,int e){ 25 ver[++tot]=v,Next[tot]=head[u],head[u]=tot,edge[tot]=e; 26 ver[++tot]=u,Next[tot]=head[v],head[v]=tot,edge[tot]=0; 27 } 28 bool bfs(){ 29 memset(dep,-1,sizeof(dep)); 30 while(!q.empty()) q.pop(); 31 for(int i=S;i<=T;++i) cur[i]=head[i]; 32 q.push(S),dep[S]=0; 33 while(!q.empty()){ 34 int u=q.front();q.pop(); 35 for(int i=head[u];i;i=Next[i]){ 36 int v=ver[i]; 37 if(dep[v]<0&&edge[i]){ 38 dep[v]=dep[u]+1,q.push(v); 39 if(v==T) return true; 40 } 41 } 42 } 43 return false; 44 } 45 int dfs(int u,int limit){ 46 if(u==T||!limit) return limit; 47 int flow=0,f; 48 for(int i=cur[u];i;cur[u]=i=Next[i]){ 49 int v=ver[i]; 50 if(dep[v]==dep[u]+1&&(f=dfs(v,min(limit,edge[i])))){ 51 flow+=f,limit-=f; 52 edge[i]-=f,edge[i^1]+=f; 53 if(!limit) break; 54 } 55 } 56 if(!flow) dep[u]=-1; 57 return flow; 58 } 59 int dinic(){ 60 int flow=0; 61 while(bfs()) flow+=dfs(S,inf); 62 return flow; 63 } 64 int main(){ 65 //freopen("testdata.in","r",stdin); 66 n=read(),m=read(),S=0,T=n+1; 67 for(int i=1,k;i<=n;++i) k=read(),k&1?add(S,i,1):add(i,T,1); 68 for(int i=1;i<=m;++i){ 69 int u=read(),v=read();add(u,v,1),add(v,u,1); 70 } 71 printf("%d ",dinic()); 72 return 0; 73 }