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.
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.
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
Recommend
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 }