BZOJ 2530 Poi2011 Party 【枚举】

 

BZOJ 2530 Poi2011 Party

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Description

Byteasar intends to throw up a party. Naturally, he would like it to be a success. Furthermore, Byteasar is quite certain that to make it so it suffices if all invited guests know each other. He is currently trying to come up with a list of his friends he would like to invite. Byteasar has friends, where is divisible by 3. Fortunately, most of Byteasar’s friends know one another. Furthermore, Byteasar recalls that he once attended a party where there were2/3 n of his friends, and where everyone knew everyone else. Unfortunately, Byteasar does not quite remember anything else from that party… In particular, he has no idea which of his friends attended it. Byteasar does not feel obliged to throw a huge party, but he would like to invite at least n/3of his friends. He has no idea how to choose them, so he asks you for help.
给定一张N（保证N是3的倍数）个节点M条边的图，并且保证该图存在一个大小至少为2N/3的团。
请输出该图的任意一个大小为N/3的团。 一个团的定义为节点的一个子集，该子集中的点两两有直接连边。 输入： 第一行是两个整数N,M。 接下来有M行，每行两个整数A,B，表示A和B有连边。保证无重边。 输出： N/3个整数，表示你找到的团。 数据范围：
3<=N<=3000,[3/2 n(2/3 n -1)]/2<=M<=[n(n-1)/2]

Input

In the first line of the standard input two integers, n and M(3<=N<=3000,[3/2 n(2/3 n -1)]/2<=M<=[n(n-1)/2]), are given, separated by a single space. These denote the number of Byteasar’s friends and the number of pairs of his friends who know each other, respectively. Byteasar’s friends are numbered from 1 to . Each of the following lines holds two integers separated by a single space. The numbers in line no.i+1(for i=1,2,…,m) are Ai and Bi(1<=Ai<Bi<=N), separated by a single space, which denote that the persons Ai and Bi now each other. Every pair of numbers appears at most once on the input.

Output

In the first and only line of the standard output your program should print N/3numbers, separated by single spaces, in increasing order. These number should specify the numbers of Byteasar’s friends whom he should invite to the party. As there are multiple solutions, pick one arbitrarily.

Sample Input

6 10
2 5
1 4
1 5
2 4
1 3
4 5
4 6
3 5
3 4
3 6

Sample Output

2 4

HINT

Explanation of the example: Byteasar’s friends numbered 1, 3, 4, 5 know one another. However, any pair of Byteasar’s friends who know each other, like 2 and 4 for instance, constitutes a correct solution, i.e., such a pair needs not be part of aforementioned quadruple.

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首先如果任意两个点之间没有连边我们把这两个点删除
因为不在团里面的点最多有n/3​，每次删除两个点，团内的点也最多删除n/3​，最少也会剩下n/3个团内的点
所以最后直接在没有被删除的点里面取n/3​个就好了

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#include<bits/stdc++.h>
using namespace std;
#define fu(a,b,c) for(int a=b;a<=c;++a)
#define N 3010
int n,m;
bool vis[N],g[N][N];
int main(){
  scanf("%d%d",&n,&m); 
  fu(i,1,m){
    int x,y;
    scanf("%d%d",&x,&y);
    g[x][y]=1;
  }
  fu(i,1,n)if(!vis[i])
    fu(j,i+1,n)if(!vis[j])
      if(!g[i][j]){vis[i]=vis[j]=1;break;}
  int siz=0;
  fu(i,1,n)if(!vis[i]&&siz<n/3)printf("%d ",i),siz++;
}