[ACM] POJ 3740 Easy Finding （DFS）

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16202		Accepted: 4349

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

Source

POJ Monthly Contest - 2009.08.23, MasterLuo

题意为：

给定由01构成的矩阵，问能不能选出几行构成新矩阵，使得新矩阵每列有且仅仅有一个1.

枚举全部行，行号递增，推断每行能否够选（是否与前面所选的行发生冲突）。当前行可选时。第j列假设为1。则用vis[j]=1标记，当行号>n(行数）时。推断每列是否都有1.

#include <iostream>
#include <stdio.h>
#include <string.h>
const int maxn=18;
const int maxm=310;
int mp[maxn][maxm];
bool vis[maxm];
int n,m;
bool yes;
using namespace std;

bool row_ok(int rth)//rth为行号，推断第rth行能够选
{
    for(int j=1;j<=m;++j)
        if(vis[j]&&mp[rth][j])//第j列已经有1了
        return false;
    for(int j=1;j<=m;++j)//能够选
        if(mp[rth][j])
            vis[j]=true;
    return true;
}

bool judge()//当选的行号大于n时。推断一下是不是每列都有1
{
    for(int j=1;j<=m;++j)
        if(!vis[j])
        return false;
    return true;
}

void dfs(int rth)
{
    if(rth>n+1)//由于当rth=n+1时，还须要推断judge()
        return;
    if(judge())//注意这两个if
    {
        yes=1;
        return;
    }
    for(int i=rth;i<=n&&!yes;++i)
    {
        if(row_ok(i))
        {
            dfs(i+1);//注意这里不是dfs(step+1)，选的行号是递增的
            for(int j=1;j<=m;++j)//还原
                if(mp[i][j])
                    vis[j]=0;
        }
    }
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=1;i<=n;++i)
            for(int j=1;j<=m;++j)
                scanf("%d",&mp[i][j]);
        memset(vis,0,sizeof(vis));
        yes=0;
        dfs(1);
        if(yes)
            printf("Yes, I found it\n");
        if(!yes)
            printf("It is impossible\n");
    }
    return 0;
}