标签：several set mem lines number return describe select blog
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 0 1 0 1 0 1 0 0
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)，求只要找到最少的顶点把所有的边都覆盖了就是最少的操作次数。
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 110 int 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; }
【二分图匹配入门专题1】D - Matrix hdu2119【最小顶点覆盖】
标签：several set mem lines number return describe select blog
原文地址：http://www.cnblogs.com/chengdongni/p/7352284.html
