POJ2692 Kindergarten 最大团 + 补图二分图

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#二分图 #最大独立集 #最大团 #补图 #图论

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本文介绍了一种解决二分图中寻找最大团问题的方法。具体地，文章通过构建补图并求解其最大独立集来间接获得最大团。文中还提供了一段C++代码示例，展示了如何实现这一算法。

Kindergarten

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7362		Accepted: 3628

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

Sample Output

Case 1: 3
Case 2: 4

Source

2008 Asia Hefei Regional Contest Online by USTC

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就是让你求二分图里的最大团，因为本身男生之间互相认识，女生之间也互相认识. 作补图之后求补图的最大独立集即可.

关于最大团的概念以及求法: Click.

#include<stdio.h>
#include<bitset>
#include<cstring>
#define clear(a) memset(a, 0, sizeof(a))
using namespace std;
const int maxn = 405;
bitset<maxn> vis;
int g, b, m, cas, num, ans;
int match[maxn], mp[maxn][maxn], h[maxn];
struct edge{ int nxt, v;} e[maxn * maxn];
inline void add(int u, int v){e[++num].v = v, e[num].nxt = h[u], h[u] = num;}
bool find(int u){
	for(int i = h[u]; i; i = e[i].nxt){
		int v = e[i].v;
		if(!vis[v]){
			vis[v] = true;
			if(!match[v] || find(match[v]))
				return match[v] = u;
		}
	}
	return false;
}
int main(){
	while(scanf("%d%d%d", &g, &b, &m), g + b + m){
		int x, y;
		for(int i = 1; i <= m; ++i) scanf("%d%d", &x, &y), mp[x][y] = true;
		for(int i = 1; i <= g; ++i)
			for(int j = 1; j <= b; ++j)
				if(!mp[i][j]) add(i, j);
		for(int i = 1; i <= g; ++i){
			if(find(i)) ans ++;
			vis = 0;
		}
		printf("Case %d: %d\n", ++cas, g + b - ans);
		
		num = ans = 0;
		clear(h), clear(mp), clear(match);
	}
}