题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2119

Matrix

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2205    Accepted Submission(s):
975

Problem Description

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.

Input

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.

Output

For 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 0

1 0 1

0 1 0

0

Sample Output

2

Author

Wendell

Source

HDU
2007-10 Programming Contest_WarmUp

题目大意：将矩阵所给的1全部变成0，每次改变可以变换一行或者一列。问最少多少次，可以使矩阵里的所有元素全部变为0？

解题思路：将横坐标X看做二分图的左支，纵坐标Y看做二分图的右支。求二分图的最小点覆盖=二分图的最匹配数。

详见代码。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4
 5 using namespace std;
 6
 7 int vis[110],Map[110][110];
 8 int ok[110],n,m;
 9
10 bool Find(int x)
11 {
12     for (int i=0;i<m;i++)
13     {
14         if (Map[x][i]==1&&!vis[i])
15         {
16             vis[i]=1;
17             if (ok[i]==-1)
18             {
19                 ok[i]=x;
20                 return true;
21             }
22             else
23             {
24                 if (Find(ok[i])==true)
25                 {
26                     ok[i]=x;
27                     return true;
28                 }
29             }
30         }
31     }
32     return false;
33 }
34
35 int main()
36 {
37     int ans;
38     while (~scanf("%d",&n))
39     {
40         ans=0;
41         if (n==0)
42             break;
43         scanf("%d",&m);
44         memset(Map,0,sizeof(Map));
45         memset(ok,-1,sizeof(ok));
46         for (int i=0;i<n;i++)
47         {
48             for (int j=0;j<m;j++)
49             {
50                 scanf("%d",&Map[i][j]);
51             }
52         }
53         for (int i=0;i<n;i++)
54         {
55             memset(vis,0,sizeof(vis));
56             if (Find(i))
57             {
58                 ans++;
59             }
60         }
61         printf ("%d\n",ans);
62     }
63     return 0;
64 }