hdu2119

Matrix

问题描述 :

Give you a matrix(only contains 0 or 1),every time you can select a row or a column and delete all the ’1′ in this row or this column .

Your task is to give out the minimum times of deleting all the ’1′ in the matrix.

输入:

There are several test cases.

The first line contains two integers n,m(1<=n,m<=100), n is the number of rows of the given matrix and m is the number of columns of the given matrix.
The next n lines describe the matrix:each line contains m integer, which may be either ‘1’ or ‘0’.

n=0 indicate the end of input.

输出:

There are several test cases.

The first line contains two integers n,m(1<=n,m<=100), n is the number of rows of the given matrix and m is the number of columns of the given matrix.
The next n lines describe the matrix:each line contains m integer, which may be either ‘1’ or ‘0’.

n=0 indicate the end of input.

样例输入:

3 3 0 0 01 0 10 1 00

样例输出:

2

题意很好理解，就是每次可以删除一行或者一列数，问最少几次可以把所有的1都变成0，也就是都删完。用二分图表示，行代表二分图的一部分，列代表二分图的一部分，map[i][j]==1代表连一条边，这样就转化为求最小顶点覆盖，即求二分图的最大匹配（边数最多的匹配，即把尽可能多的边与某一个顶点相关联，这样选择全部的边所需要的最少顶点就是最小顶点覆盖）。此外，二分图还有最小路径覆盖,意思是用最少的边把图中所有的顶点都遍历到（最小路径覆盖 = 顶点数 – 最大匹配）。

#include<iostream> #include<cstring> const int N=110; using namespace std; int map[N][N]; int visited[N]; int used[N]; int n,m;  //从定点x出发，用深度优先搜索策略寻找增广路 int Solve(int x){     for(int j=0;j<m;j++){         if(map[x][j]&&!visited[j]){             visited[j]=1;             //如果没有匹配，或者已经匹配了，但从used[j]出发可以找到一天增广路；             //如果前一个条件成立，则不会递归调用             if(used[j]==-1||Solve(used[j])){                 used[j]=x;                 return 1;             }         }     }     return 0; }   //求二部图最大匹配的匈牙利算法 int MaxMatch(){     int count=0;     for(int i=0;i<n;i++){         memset(visited,0,sizeof(visited));         if(Solve(i))count++;     }     return count; }  int main(){     while(~scanf("%d",&n)&&n){         scanf("%d",&m);         for(int i=0;i<n;i++){             for(int j=0;j<m;j++){                 scanf("%d",&map[i][j]);             }         }         memset(used,-1,sizeof(used));         int ans=MaxMatch();         printf("%d\n",ans);     }     return 0; }