HNU 10111 0-1矩阵

http://acm.hnu.cn/online/?action=problem&type=show&id=10111

题意：中文

题解：在龙哥的帮助下正了二分图匹配的三观……以前的理解繁琐，或者有点儿错吧……二分图匹配从左往右匹配，找增广路。顶点数和match（）不需要那么麻烦。

//
//  main.cpp
//  POJ 3041
//
//  Created by zhang on 14-4-16.
//  Copyright (c) 2014年 apple. All rights reserved.
//

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <string>
#include <vector>
#include <list>
#include <map>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <numeric>
#include <functional>
#include <set>
#include <fstream>

using namespace std;

const int maxn=10010;
int g1[maxn];
int g2[maxn];
int V=0;
vector<int> G[maxn*2 ];
int match[maxn];
bool used[maxn];
int g[maxn][maxn];

void add_edge(int u,int v){
    G[u].push_back(v);
    G[v].push_back(u);
}

bool dfs(int v)
{
    used[v]=true;
    for (int i=0; i<G[v].size(); i++) {
        int u=G[v][i];
        int w=match[u];
        if(used[w]) continue;
        if (w<0||dfs(w)) {
           // match[v]=u;
            match[u]=v;
            return true;
        }
    }
    return false;
}

int b_match()
{
    int res=0;
    memset(match, -1, sizeof(match));
    for (int v=0; v<V; v++) {
        if (match[v]<0) {
            memset(used, 0, sizeof(used));
            if (dfs(v)) {
                res++;
            }
        }
    }
    return res;
}
int main()
{
    //freopen("/Users/apple/Desktop/HNU 10111/HNU 10111/in", "r", stdin);
    //freopen("/Users/apple/Desktop/HNU 10111/HNU 10111/out", "w", stdout);
    int N,M;
    char s[110][110];
    while((scanf("%d",&N))!=EOF&&N!=0){
        scanf("%d",&M);
        V=N;
        //cout<<V<<endl;
        int K=0;
        int p=0;
        int q=0;
        getchar();
        for (int i=0; i<N; i++) {
            for (int j=0; j<M; j++) {
                s[i][j]=getchar();
              //  cout<<s[i][j];
                if (s[i][j]=='1') {
                    K++;
                    g1[p]=i;
                    g2[q]=j;
                    //cout<<"g1 g2 "<<i <<" "<<j<<" "<<g1[p]<<" "<<g2[q]<<endl;
                    p++;
                    q++;
                }
            }
            getchar();
           // puts("");
        }
        //cout<<"K "<<K<<endl;
        for (int i=0; i<K; i++) {
            //cout<<g1[i]<<g2[i]<<endl;
            add_edge(g1[i],N +g2[i]);
         }
        printf("%d
",b_match());
        for (int i=0; i<K*2; i++) {
            G[i].clear();
        }
    }
    
    return 0;
}