题目链接：http://poj.org/problem?id=3692
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Kindergarten
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 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 Source |
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这个题目是二分匹配模板题，一开始用浙大模板里的最大团模板搞，超时了。。。（至今不明白为什么）
后来利用
最大团=原图节点数-补图的最大二分匹配数
AC代码：
//最大团问题，可以用原图节点数减去其补图的二分匹配
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<vector>
#include<string>
#include<cmath>
#include<set>
#include<map>
#include<cstdlib>
#define CLR(A) memset(A,0,sizeof(A))
using namespace std;
const int MAX = 210;
int b,g,m,link[MAX];
bool mp[MAX][MAX], vis[MAX];
bool dfs(int u){
for(int i = 1; i <= b; i ++)
if(!vis[i] && !mp[u][i]){
vis[i] = true;
if(link[i] == -1 || dfs(link[i])){
link[i] = u;
return true;
}
}
return false;
}
int main(){
int ncase = 0;
while(~scanf("%d%d%d",&g,&b,&m)&&g+b+m){
CLR(mp);
while(m --){
int u, v;
scanf("%d%d",&u,&v);
mp[u][v]=1;
}
int t=0;
memset(link, -1, sizeof(link));
for(int i = 1; i <= g; i ++){
memset(vis, 0, sizeof(vis));
if(dfs(i)) t++;
}
cout<<"Case "<<++ncase<<": "<<g+b-t<< endl;
}
return 0;
}
TLE代码：
#include<iostream>
#include<vector>
#include<cstring>
#include<cstdio>
using namespace std;
#define V 600
int g[V][V],dp[V],stk[V][V],mx;
int dfs(int n,int ns,int dep){
if(0==ns){
if(dep>mx) mx=dep;
return 1;
}
int i,j,k,p,cnt;
for(i=0;i<ns;i++){
k=stk[dep][i];cnt=0;
if(dep+n-k<=mx) return 0;
if(dep+dp[k]<=mx) return 0;
for(j=i+1;j<ns;j++){
p=stk[dep][j];
if(g[k][p]) stk[dep+1][cnt++]=p;
}
dfs(n,cnt,dep+1);
}
return 1;
}
int clique(int n){
int i,j,ns;
for(mx=0,i=n-1;i>=0;i--){
for(ns=0,j=i+1;j<n;j++){
if(g[i][j]) stk[1][ns++]=j;
dfs(n,ns,1); dp[i]=mx;
}
}
return mx;
}
int main(){
int G,B,M,kase=0;
while(~scanf("%d%d%d",&G,&B,&M) && (G || B || M)){
memset(g,0,sizeof(g));
for(int i=0;i<G;i++)
for(int j=0;j<G;j++){
// if(i==j) continue;
g[i][j]=1;
}
for(int i=0;i<B;i++)
for(int j=0;j<B;j++){
// if(i==j) continue;
g[i+G][j+G]=1;
}
for(int i=0;i<M;i++){
int x,y;
scanf("%d%d",&x,&y);
x--;y--;
g[x][y+G]=1;
g[y+G][x]=1;
}
int res=clique(G+B);
printf("Case %d: %d\n",++kase,res);
}
}
poj 3692 Kindergarten （最大团模板题）,布布扣,bubuko.com
poj 3692 Kindergarten （最大团模板题）
原文地址：http://blog.csdn.net/asdfghjkl1993/article/details/38711177
