HDU--3605 Escape (多重匹配)

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 1
1
1

2 2
1 0
1 0
1 1

Sample Output

YES
NO

思路：用一个二维数组存与星球匹配的人员的编号，若有人无法匹配，则输出NO
AC代码：

#include<cstdio>
#include<cstring>
const int N=100010;
bool vis[15];
int map
[15],mat[15]
;
int cnt[15],vol[15];
int n,m;
bool find(int x)
{
int i,j;
for(i=0;i<m;i++)
{
if(!vis[i]&&map[x][i])
{
vis[i]=true;
if(cnt[i]<vol[i])
{
mat[i][cnt[i]++]=x;
return true;
}
for(j=0;j<cnt[i];j++)
{
if(find(mat[i][j]))
{
mat[i][j]=x;
return true;
}
}
}
}
return false;
}
int MMG()
{
int i;
memset(cnt,0,sizeof(cnt));
for(i=0;i<n;i++)
{
memset(vis,false,sizeof(vis));
if(!find(i)) return false;
}
return true;
}
int main()
{
int i,j;
while(scanf("%d%d",&n,&m)!=EOF)
{
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",&vol[i]);
if(MMG()) printf("YES\n");
else printf("NO\n");
}
return 0;
}