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 isM, N ( M ≤ 16, N ≤ 300). The next M lines every line containsN 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 <algorithm>
using namespace std;
const int maxn=500;
bool mark[maxn];
int vis[maxn][maxn];
int m,n;
bool ans;
void dfs(int c)
{
	int count=0;
	for(int i=1;i<=m;i++)
	{
		if(mark[i]&&vis[i][c])
			count++;
	}
	if(count>1) return;
	if(count==1)
	{
		if(c<n) dfs(c+1);
		else
		{
			ans=true;
			return ;
		}
	}
	if(count==0)
	{
		for(int i=1;i<=m;i++)
		{
			bool work=true;
			if(!mark[i]&&vis[i][c])
			{
				for(int j=1;j<c;j++)
				{
					if(vis[i][j])
					{
						work=false;
						break;
					}
				}
				if(work)
				{
					mark[i]=true;
					if(c<n) dfs(c+1);
					else ans=true;
					mark[i]=false;
				}
			}
		}
	}
}
int main()
{
	int a;
	while(scanf("%d%d",&m,&n)!=EOF)
	{
		memset(vis,0,sizeof(vis));
		memset(mark,0,sizeof(mark));
		for(int i=1;i<=m;i++)
		{
			for(int j=1;j<=n;j++)
			{
				scanf("%d",&a);
				vis[i][j]=a==1;
			}
		}
		ans=false;
		dfs(1);
		if(ans) printf("Yes, I found it\n");
		else printf("It is impossible\n");
	}
	return 0;
}