HDU——T 2119 Matrix

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

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3097 Accepted Submission(s): 1429

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

Author

Wendell

Source

HDU 2007-10 Programming Contest_WarmUp

Recommend

威士忌 | We have carefully selected several similar problems for you: 2115 2117 2114 2118 2120

格点为一的格子坐标连边

 1 #include <cstring>
 2 #include <cstdio>
 3 
 4 bool vis[110];
 5 int map[110][110];
 6 int n,m,match[110];
 7 
 8 bool find(int x)
 9 {
10     for(int y=1;y<=m;y++)
11         if(map[x][y]&&!vis[y])
12         {
13             vis[y]=1;
14             if(!match[y]||find(match[y]))
15             {
16                 match[y]=x;
17                 return true;
18             }
19         }
20     return false;
21 }
22 
23 int main()
24 {
25     for(;scanf("%d",&n)&&n;)
26     {
27         int ans=0;
28         scanf("%d",&m);
29         for(int i=1;i<=n;i++)
30             for(int j=1;j<=m;j++)
31             scanf("%d",&map[i][j]);
32         for(int i=1;i<=n;i++)
33           {
34               if(find(i)) ans++;
35               memset(vis,0,sizeof(vis));
36           }
37         printf("%d
",ans);
38         memset(map,0,sizeof(map));
39         memset(match,0,sizeof(match));
40     }
41     return 0;
42 }

——每当你想要放弃的时候，就想想是为了什么才一路坚持到现在。