poj 3692 Kindergarten

Kindergarten

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 6956		Accepted: 3436

Description

In a kindergarten, there are a lot of kids. All girls of the kids know each other and all boys also know each other. In addition to that, some girls and boys know each other. Now the teachers want to pick some kids to play a game, which need that all players know each other. You are to help to find maximum number of kids the teacher can pick.

Input

The input consists of multiple test cases. Each test case starts with a line containing three integers
G, B (1 ≤ G, B ≤ 200) and M (0 ≤ M ≤ G × B), which is the number of girls, the number of boys and
the number of pairs of girl and boy who know each other, respectively.
Each of the following M lines contains two integers X and Y (1 ≤ X≤ G,1 ≤ Y ≤ B), which indicates that girl X and boy Y know each other.
The girls are numbered from 1 to G and the boys are numbered from 1 to B.

The last test case is followed by a line containing three zeros.

Output

For each test case, print a line containing the test case number( beginning with 1) followed by a integer which is the maximum number of kids the teacher can pick.

Sample Input

2 3 3
1 1
1 2
2 3
2 3 5
1 1
1 2
2 1
2 2
2 3
0 0 0

Sample Output

Case 1: 3
Case 2: 4

翻译：小朋友手拉手做游戏，现在老师希望从小朋友当中挑出一个集合，集合中的小朋友全都认识，那么这个集合最大是多少。
思路：显然是个最大独立集问题，只是男女小朋友关系图的补图的最大独立集，其补图中两个人相互连线意味着相互不认识，从图中挑出来的最大独立集就是题设要求的。
ＡＣ代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<queue>
#include<set>
#include<vector>
#include<cstring>
#include<string>
using namespace std;
#define INF 0x3f3f3f3f
const int N_MAX = 500;
int V;//点的个数
vector<int>G[N_MAX];
int match[N_MAX];
bool used[N_MAX];
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], w = match[u];
        if (w < 0 || !used[w] && dfs(w)) {
            match[v] = u;
            match[u] = v;
            return true;
        }
    }
    return false;
}

int bipartite_matching() {
    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;
}
bool vis[N_MAX][N_MAX];
int g, B, M;
int main() {
    int Case = 0;
    while (scanf("%d%d%d",&g,&B,&M)&&g) {
        Case++;
        //0~g-1:girl
        //g~g+B-1:boy
        V = g + B;
        memset(vis,0,sizeof(vis));
        for (int i = 0; i < M; i++) {
            int a, b;
            scanf("%d%d",&a,&b);
            a--, b--;
            vis[a][g + b]=true;
        }
        for (int i=0; i<g;i++) {
            for (int j = g; j < V;j++) {
                if (!vis[i][j]) {//建立原图的补图
                    add_edge(i,j);
                }
            }
        }
        printf("Case %d: %d
",Case,V-bipartite_matching());
        for (int i = 0; i < V;i++) {
            G[i].clear();
        }
    }
    return 0;
}