容易想到网络流。将每个人拆成0和1两个点。若某人值为0的话则让源连向0，否则让1连向汇，流量为1。相互认识的人之间01各自连边。跑最小割即可。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; int read() { int x=0,f=1;char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } #define N 310 #define S 0 #define T 601 int n,m,p[N<<1],t=-1,ans=0; int d[N<<1],cur[N<<1],q[N<<1]; struct data{int to,nxt,cap,flow; }edge[N*N<<1]; int trans(int x,int y){return (x<<1)-y;} void addedge(int x,int y) { t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].cap=1,edge[t].flow=0,p[x]=t; t++;edge[t].to=x,edge[t].nxt=p[y],edge[t].cap=0,edge[t].flow=0,p[y]=t; } bool bfs() { memset(d,255,sizeof(d));d[S]=0; int head=0,tail=1;q[1]=S; do { int x=q[++head]; for (int i=p[x];~i;i=edge[i].nxt) if (d[edge[i].to]==-1&&edge[i].flow<edge[i].cap) { d[edge[i].to]=d[x]+1; q[++tail]=edge[i].to; } }while (head<tail); return ~d[T]; } int work(int k,int f) { if (k==T) return f; int used=0; for (int i=cur[k];~i;i=edge[i].nxt) if (d[k]+1==d[edge[i].to]) { int w=work(edge[i].to,min(f-used,edge[i].cap-edge[i].flow)); edge[i].flow+=w,edge[i^1].flow-=w; if (edge[i].flow<edge[i].cap) cur[k]=i; used+=w;if (used==f) return f; } if (used==0) d[k]=-1; return used; } void dinic() { while (bfs()) { memcpy(cur,p,sizeof(p)); ans+=work(S,N*N); } } int main() { #ifndef ONLINE_JUDGE freopen("bzoj2768.in","r",stdin); freopen("bzoj2768.out","w",stdout); const char LL[]="%I64d\n"; #else const char LL[]="%lld\n"; #endif n=read(),m=read(); memset(p,255,sizeof(p)); for (int i=1;i<=n;i++) { int x=read(); if (x) addedge(trans(i,1),T); else addedge(S,trans(i,0)); } for (int i=1;i<=m;i++) { int x=read(),y=read(); addedge(trans(x,0),trans(y,1)); addedge(trans(y,0),trans(x,1)); } dinic(); cout<<ans; return 0; }