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POJ 3740 Easy Finding(dfs回溯)

2024-11-20 04:31:39   203人阅读

B - Easy Finding
Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u
Submit Status Practice POJ 3740

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 Nintegers 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

这是一道子集枚举的回溯题，针对当前行选不选进行回溯，并且进行剪枝。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define maxn 1005
int n,m;
int a[20][305];
int vis[305];
bool judge(){
   for(int i=0;i<n;i++)
    if(vis[i]==0)return false;
   return true;
}
bool judges(){
   for(int i=0;i<n;i++)
    if(vis[i]>1)return false;
   return true;
}
bool dfs(int cur)
{
    if(judge())return true;
    if(cur>=m)return judge();
    for(int i=0;i<n;i++){
        if(a[cur][i]==1)vis[i]++;
    }
    if(judges()){
        if(dfs(cur+1))return true;
    }
    for(int i=0;i<n;i++){
        if(a[cur][i]==1)vis[i]--;
    }
    if(dfs(cur+1))return true;
    return false;
}
int main()
{
    //freopen("in.txt","r",stdin);
    while(scanf("%d%d",&m,&n)!=EOF){
        for(int i=0;i<m;i++)
            for(int j=0;j<n;j++)
            scanf("%d",&a[i][j]);
        memset(vis,0,sizeof vis);
        if(dfs(0))puts("Yes, I found it");
        else puts("It is impossible");

    }

}


POJ 3740 Easy Finding(dfs回溯)


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