hdu 3605 Escape（多重匹配）

#include<stdio.h>#include<string.h>#define N 100005#define M 13bool map[N][M],vis[M];int m,n,match[M][N],lim[M],num[M];//match[星球][哪个人]=多少人。num[哪个星球]=目前有多少人int find(int u){for(int i=0;i<m;i++){if(!vis[i]&&map[u][i]){vis[i]=1;if(num[i]<lim[i]){match[i][num[i]++]=u;return 1;}for(int j=0;j<num[i];j++)if(find(match[i][j])){match[i][j]=u;return 1;}}}return 0;}int hungary(){for(int i=0;i<n;i++){memset(vis,0,sizeof(vis));if(!find(i))break; //不能写成 sum+=find(i);会超时的。}return i;}int main(){int i,j;while(scanf("%d%d",&n,&m)!=-1){memset(match,0,sizeof(match));memset(map,0,sizeof(map));memset(num,0,sizeof(num));for(i=0;i<n;i++){for(j=0;j<m;j++)scanf("%d",&map[i][j]);}for(i=0;i<m;i++) scanf("%d",&lim[i]);if(hungary()==n)printf("YES\n");else printf("NO\n");}return 0;}