题目要的并不是最大匹配下得到的最大的结果。
网上流行的做法是加边。其实，在连续增广的时候求得一个可行流的总费用为负就停止这样就行了。
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
#define INF (1<<30)
#define MAXN 222
#define MAXM 22222
struct Edge{
int u,v,cap,cost,next;
}edge[MAXM];
int head[MAXN];
int NV,NE,vs,vt; void addEdge(int u,int v,int cap,int cost){
edge[NE].u=u; edge[NE].v=v; edge[NE].cap=cap; edge[NE].cost=cost;
edge[NE].next=head[u]; head[u]=NE++;
edge[NE].u=v; edge[NE].v=u; edge[NE].cap=; edge[NE].cost=-cost;
edge[NE].next=head[v]; head[v]=NE++;
}
bool vis[MAXN];
int d[MAXN],pre[MAXN];
bool SPFA(){
for(int i=;i<NV;++i){
vis[i]=;
d[i]=INF;
}
vis[vs]=;
d[vs]=;
queue<int> que;
que.push(vs);
while(!que.empty()){
int u=que.front(); que.pop();
for(int i=head[u]; i!=-; i=edge[i].next){
int v=edge[i].v;
if(edge[i].cap && d[v]>d[u]+edge[i].cost){
d[v]=d[u]+edge[i].cost;
pre[v]=i;
if(!vis[v]){
vis[v]=;
que.push(v);
}
}
}
vis[u]=;
}
return d[vt]!=INF;
}
int MCMF(){
int res=;
while(SPFA()){
int flow=INF,cost=;
for(int u=vt; u!=vs; u=edge[pre[u]].u){
flow=min(flow,edge[pre[u]].cap);
}
for(int u=vt; u!=vs; u=edge[pre[u]].u){
edge[pre[u]].cap-=flow;
edge[pre[u]^].cap+=flow;
cost+=flow*edge[pre[u]].cost;
}
if(cost>=) break;
res+=cost;
}
return res;
}
int main(){
int n,a[],b;
while(~scanf("%d",&n)&&n){
for(int i=;i<=n;++i) scanf("%d",a+i);
vs=; vt=n<<|; NV=vt+; NE=;
memset(head,-,sizeof(head));
for(int i=;i<=n;++i){
addEdge(vs,i,,);
addEdge(i+n,vt,,);
}
for(int i=;i<=n;++i){
for(int j=;j<=n;++j){
scanf("%1d",&b);
if(b) addEdge(i,j+n,,-(a[i]^a[j]));
}
}
printf("%d\n",-MCMF());
}
return ;
}