hdu 3605 Escape（多重匹配）

Escape

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2836    Accepted Submission(s): 773

Problem Description

2012 If this is the end of the world how to do? I do not know how. But now scientists have found that some stars, who can live, but some people do not fit to live some of the planet. Now scientists want your help, is to determine what all of people can live in these planets.

 

Input

More set of test data, the beginning of each data is n (1 <= n <= 100000), m (1 <= m <= 10) n indicate there n people on the earth, m representatives m planet, planet and people labels are from 0. Here are n lines, each line represents a suitable living conditions of people, each row has m digits, the ith digits is 1, said that a person is fit to live in the ith-planet, or is 0 for this person is not suitable for living in the ith planet.
The last line has m digits, the ith digit ai indicates the ith planet can contain ai people most..
0 <= ai <= 100000

 

Output

Determine whether all people can live up to these stars
If you can output YES, otherwise output NO.

 

Sample Input

1 1112 21 01 01 1

 

Sample Output

YESNO
题意：给出n、m，表示有n个人和m个星球。然后是n行，每行有m个数字，第i行第j个数字为1时表示编号为i的人可以适应在j星球生活。最后是一行含有m个数，表示每个星球的可以容纳的人口数。
思路：二分图的多重匹配。
AC代码：
#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <vector>#include <cmath>#include <stack>#include <cstdlib>using namespace std;const int maxn=100005;const int maxm=15;int match[maxm][maxn],vol[maxm],cnt[maxm];bool map[maxn][maxm],vis[maxm];int n,m;bool find(int x){    for(int i=1;i<=m;i++)    if(!vis[i]&&map[x][i])    {        vis[i]=true;        if(cnt[i]<vol[i])        {        match[i][cnt[i]++]=x;        return true;        }        for(int j=0;j<cnt[i];j++)        if(find(match[i][j]))        {            match[i][j]=x;            return true;        }    }    return false;}bool MMG(){    memset(cnt,0,sizeof(cnt));    for(int i=1;i<=n;i++)    {        memset(vis,false,sizeof(vis));        if(!find(i)) return false;    }    return true;}int main(){   while(scanf("%d%d",&n,&m)!=EOF)   {       for(int i=1;i<=n;i++)       for(int j=1;j<=m;j++)       scanf("%d",&map[i][j]);       for(int i=1;i<=m;i++)       scanf("%d",&vol[i]);       if(MMG())       printf("YES\n");       else       printf("NO\n");   }   return 0;}