(水 dfs) poj 3740

Easy Finding

Description

Given a M×N matrix A. A_ij ∈ {0, 1} (0 ≤ i < M, 0 ≤ j < N), could you find some rows that let every cloumn contains and only contains one 1.

Input

There are multiple cases ended by EOF. Test case up to 500.The first line of input is M, N (M ≤ 16, N ≤ 300). The next M lines every line contains N integers separated by space.

Output

For each test case, if you could find it output "Yes, I found it", otherwise output "It is impossible" per line.

Sample Input

3 3
0 1 0
0 0 1
1 0 0
4 4
0 0 0 1
1 0 0 0
1 1 0 1
0 1 0 0

Sample Output

Yes, I found it
It is impossible

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<queue>
#include<vector>
#include<stack>
using namespace std;
int n,m,mp[305][305],vis[305];
bool check()
{
    for(int i=0;i<m;i++)
        if(vis[i]!=1)
            return false;
    return true;
}
bool ok()
{
    for(int i=0;i<m;i++)
    {
        if(vis[i]>1)
            return false;
    }
    return true;
}
bool dfs(int x)
{
    if(check())
        return true;
    for(int i=x;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            vis[j]+=mp[i][j];
        }
        if(ok())
        {
            if(dfs(i+1))
                return true;
        }
        for(int j=0;j<m;j++)
        {
            vis[j]-=mp[i][j];
        }
    }
    return false;
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(vis,0,sizeof(vis));
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
                scanf("%d",&mp[i][j]);
        }
        if(dfs(0))
        {
            printf("Yes, I found it\n");
        }
        else
            printf("It is impossible\n");
    }
    return 0;
}