2014
11-27

# Escape

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4298    Accepted Submission(s): 1129

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

Source

X

#include"stdio.h"
#include"string.h"
#define N 100005
#define M 15
bool g[N][M],vis[M];          //存边的权值0、1，记录是否访问过
int n,m;
int find(int k)
{
int i,j;
for(i=0;i<m;i++)
{
if(!vis[i]&&g[k][i])
{
vis[i]=1;
if(cnt[i]<lim[i])
{
return 1;
}
for(j=0;j<cnt[i];j++)
{
{
return 1;
}
}
}
}
return 0;
}
int main()
{
int i,j;
while(scanf("%d%d",&n,&m)!=-1)
{
memset(g,0,sizeof(g));
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
{
scanf("%d",&g[i][j]);
}
}
for(i=0;i<m;i++)
{
scanf("%d",&lim[i]);
}
memset(cnt,0,sizeof(cnt));
for(i=0;i<n;i++)
{
memset(vis,0,sizeof(vis));
if(!find(i))
break;
}
if(i==n)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}