【二分图匹配入门专题1】D

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.

InputThere 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. 
OutputFor each of the test cases, in the order given in the input, print one line containing the minimum times of deleting all the '1' in the matrix. 
Sample Input

3 3 0 0 01 0 10 1 00

Sample Output

2题意：输入n行m列的数，这些数只有1或0组成，每次可以消掉一行或者一列的1，问，消灭掉全部的1最少需要进行多少次操作思路：我二分图匹配模型还是没有转换过来，但是队友给我讲了以后，豁然开朗，矩阵的行x就相当于左集合，矩阵的列y就相当于右集合，要使1被消灭，就相当于连通(x,y)，求最少需要多少次操作可以连通所有1的(x,y)这不就是最小顶点覆盖吗。sum没有初始化样例也能过，但是wrong了三次后才发现是这个原因，尴尬~~

#include<stdio.h>#include<string.h>#define N 110int book[N],e[N][N],match[N];int n,m;int dfs(int u){    int i;    for(i = 1; i <= m; i ++)    {        if(!book[i]&&e[u][i])        {            book[i] = 1;            if(!match[i]||dfs(match[i]))            {                match[i] = u;                return 1;            }        }    }    return 0;}int main(){    int i,j,sum;    while(scanf("%d",&n),n!=0)    {        scanf("%d",&m);        memset(match,0,sizeof(match));        memset(e,-1,sizeof(e));        for(i = 1; i <= n; i ++)            for(j = 1; j <= m; j ++)                scanf("%d",&e[i][j]);        sum = 0;        for(i = 1; i <= n; i ++)        {            memset(book,0,sizeof(book));            if(dfs(i))                sum ++;        }        printf("%d\n",sum);    }    return 0;}