Matrix
http://acm.hdu.edu.cn/showproblem.php?pid=2119
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
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
#include<cstdio> #include<cstring> using namespace std; int n,m,a[101][101]; int tot,front[101],nxt[10010],to[10010]; int match[101],ans; bool v[101]; void add(int u,int v) { to[++tot]=v; nxt[tot]=front[u]; front[u]=tot; } bool go(int u) { for(int i=front[u];i;i=nxt[i]) { if(!v[to[i]]) { v[to[i]]=true; if(!match[to[i]]||go(match[to[i]])) { match[to[i]]=u; return 1; } } } return 0; } int main() { while(scanf("%d",&n)) { if(!n) return 0; scanf("%d",&m); ans=0; tot=0; memset(front,0,sizeof(front)); memset(match,0,sizeof(match)); for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) { scanf("%d",&a[i][j]); if(a[i][j]) add(i,j); } for(int i=1;i<=n;i++) { memset(v,0,sizeof(v)); if(go(i)) ans++; } printf("%d ",ans); } }