hdu2119—Matrix（最小点覆盖）

题目链接：传送门

Matrix

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3012    Accepted Submission(s): 1368

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 01 0 10 1 00

 

Sample Output

2

 

Author

Wendell

 

解题思路：将每一行看成一个点ri，将每一列看成一个点ci，在‘1’存在的位置对ri和cj连边，即表示第i行第j列有一个1，最后得到一个二分图。若想要删掉矩形中的所有1，只要删掉二分图的所有边即可，删掉图中的某一个点相当于删掉矩形中的某一行或某一列，最少删掉多少个点能把二分图中的所有边删除，即变成了最小点覆盖问题，即对这个二分图求个最大匹配

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <string>#include <stack>#include <queue>using namespace std;typedef long long ll;const int N = 10900;const int M = 109;const int INF = 0x3fffffff;const double eps = 1e-8;const double PI = acos(-1.0);struct Edge{    int node;    Edge*next;}m_edge[N];int girl[M];Edge*head[M];int Flag[M],Ecnt,cnt;void init(){    Ecnt = cnt = 0;    fill( girl , girl+M , 0 );    fill( head , head+M , (Edge*)0 );}//b对g有好感void mkEdge( int b , int g ){    m_edge[Ecnt].node = g;    m_edge[Ecnt].next = head[b];    head[b] = m_edge+Ecnt++;}bool find( int x ){    for( Edge*p = head[x] ; p ; p = p->next ){        int s = p->node;   //有好感的女生        if( !Flag[s] ){            Flag[s] = true; //该女生在本轮匹配中被访问            if( girl[s] == 0 || find(girl[s]) ){                //女生没有对象或者另外一个男生能把这个妹纸让给x男                girl[s] = x;                return true;            }        }    }    return false;}//构建二分图void Build( int n , int m ){    int r;    for( int i = 1 ; i <= n ; ++i ){        for( int j = 1 ; j <= m ; ++j ){            scanf("%d",&r);            if( r == 1 ) mkEdge(i,j);        }    }}void solve( int n ){    for( int i =  1 ; i <= n ; ++i ){        fill( Flag , Flag+M , 0 );        if( find(i) ) ++cnt;    }}int main(){    int n,m;    while( ~scanf("%d",&n)&&n ){        scanf("%d",&m);        init();        Build(n,m);        solve(n);        printf("%d\n",cnt);    }    return 0;}