[POJ3740] Easy Finding

Description

Given a M×N matrix A. \(Aij ∈ {0, 1} (0 ≤ i < M, 0 ≤ j < N)\), 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

Source

POJ Monthly Contest - 2009.08.23, MasterLuo

题解

由于行\(≤16\)，所以我们可以把行压缩成二进制，用\(l[]\)记录每一列的\(1\)出现在那几行，如果有一个\(l[i]\)为空，则不存在解，为了保证每一列的唯一性，我们可以发现，当每一列只有与之匹配的一行时，状态才为\(2^n\)，判断一个数\(x\)是否为\(2^n\)，只需判断\(x\)&\((x-1)\)即可。

#include<iostream>
#include<cstdio>
using namespace std;

const int N=18,M=301;
int n,m,two[N],l[M];

bool Can()
{
	for(int i=1;i<=m;++i)
		if(!l[i]) return 0;
	return 1;
}

bool Chk(int nw)
{
	int YH;
	for(int i=1;i<=m;++i)
	{
		YH=nw&l[i];
		if(YH&(YH-1)) return 0;
	}
	return 1;
}

bool Dfs(int id,int nw)//id表示现在是第几列，nw记录用了哪些行
{
	if(id>m) return 1;
	if(nw&l[id]) return Dfs(id+1,nw);
	for(int i=1;i<=n;++i)
		if((l[id]&two[i])&&Chk(nw|two[i])&&Dfs(id+1,nw|two[i])) return 1;
	return 0;
}

int main()
{
	int tmp; bool Ans;
	two[0]=1; for(int i=1;i<N;++i) two[i]=two[i-1]<<1;
	while(~scanf("%d%d",&n,&m))
	{
		for(int i=1;i<=m;++i) l[i]=0;
		for(int i=1;i<=n;++i)
			for(int j=1;j<=m;++j)
				if(scanf("%d",&tmp)&&tmp) l[j]|=two[i];
		Ans=0;
		if(Can()) Ans=Dfs(1,0);
		if(Ans) puts("Yes, I found it");
		else puts("It is impossible");
	}
	return 0;
}