poj 3692 Kindergarten 【最大团 = 补图最大独立集 = 补图节点数 - 补图最大匹配】

Kindergarten

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 5922 Accepted: 2899

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 31 11 22 32 3 51 11 22 12 22 30 0 0

Sample Output

Case 1: 3Case 2: 4

题意：给定一个无向图，让你求它的最大团。

思路：最大团 = 补图最大独立集 = 补图节点数 - 补图最大匹配。

因为B、G弄反了，WA2次。。。

AC代码：

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <vector>#define INF 0x3f3f3f3f#define eps 1e-8#define MAXN (200+10)#define MAXM (50000000)#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while(a--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1using namespace std;int B, G, M;int Map[MAXN][MAXN];bool used[MAXN];int match[MAXN];int DFS(int u){    for(int i = 1; i <= B; i++)    {        if(!used[i] && Map[u][i] == 0)        {            used[i] = true;            if(match[i] == -1 || DFS(match[i]))            {                match[i] = u;                return 1;            }        }    }    return 0;}int main(){    int kcase = 1;    while(scanf("%d%d%d", &G, &B, &M) != EOF)    {        if(B == 0 && G == 0 && M == 0)            break;        CLR(Map, 0);        for(int i = 1; i <= M; i++)        {            int a, b;            Ri(a); Ri(b);            Map[a][b] = 1;        }        CLR(match, -1);        int ans = 0;        for(int i = 1; i <= G; i++)        {            CLR(used, false);            ans += DFS(i);        }        printf("Case %d: %d\n", kcase++, B+G-ans);    }    return 0;}