zoj3646二分匹配

这是我zoj的第一道题，比赛的时候怎么也想到是二分匹配。想到将它转化为图，将U看做是1，D 看做是0，将整张图看做是一个邻接矩阵，求每个点是否有自环或者是回路。但是，这都没想到是二分匹配。。。。惭愧啊。。。。

#include <iostream>#include <stdio.h>using namespace std;const int oo=1e8;const int mm=111111;const int mn=999;int node,src,dest,edge;int ver[mm],flow[mm],next[mm];int head[mn],work[mn],dis[mn],q[mn];void prepare(int _node,int _src,int _dest){    node=_node,src=_src,dest=_dest;    for(int i=0;i<node;i++) head[i]=-1;    edge=0;}void addedge(int u,int v,int c){    ver[edge]=v;flow[edge]=c;next[edge]=head[u];head[u]=edge++;    ver[edge]=u;flow[edge]=0;next[edge]=head[v];head[v]=edge++;}bool Dinic_bfs(){    int i,u,v,l,r=0;    for(i=0;i<node;i++)        dis[i]=-1;    dis[q[r++]=src]=0;    for(l=0;l<r;l++)        for(i=head[u=q[l]];i>=0;i=next[i])            if(flow[i]&&dis[v=ver[i]]<0)            {                dis[q[r++]=v]=dis[u]+1;                if(v==dest) return 1;            }    return 0;}int Dinic_dfs(int u,int exp){    if(u==dest) return exp;    for(int &i=work[u],v,tmp;i>=0;i=next[i])    {        if(flow[i]&&dis[v=ver[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0)        {            flow[i]-=tmp;            flow[i^1]+=tmp;            return tmp;        }    }    return 0;}int Dinic_flow(){    int i,ret=0,delta;    while(Dinic_bfs())    {        for(i=0;i<node;i++)            work[i]=head[i];        while(delta=Dinic_dfs(src,oo))  ret+=delta;    }    return ret;}int main(){    int n,tmpx,i,j;    char ch[mn];    while(scanf("%d",&n)!=EOF)    {       prepare(n+n+2,0,n+n+1);        for(i=1;i<=n;i++)        {            scanf("%s",ch);            for(j=0;j<n;j++)            {                if(ch[j]=='U')  addedge(i,n+j+1,1);            }        }        for(i=1;i<=n;i++)           addedge(src,i,1),addedge(i+n,dest,1);        tmpx=Dinic_flow();        if(tmpx==n) printf("YES\n");        else printf("NO\n");    }    return 0;}