BZOJ2530 [Poi2011]Party 【贪心】

题目链接

BZOJ2530

题解

如果我们删去一对不连边的仍然存在的点的话，这对点肯定不同时在那个\(\frac{2}{3}n\)的团中，也就是说，每次删点至少删掉一个外点，至多删掉一个内点

那么我们要删掉团外的点最多使用\(\frac{1}{3}n\)个团内的点就可以了，剩下的至少\(\frac{1}{3}n\)个点就在一个团内

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 3005,maxm = 100005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
int G[maxn][maxn],vis[maxn],n,m;
int main(){
	n = read(); m = read(); int a,b;
	while (m--){
		a = read(); b = read();
		G[a][b] = G[b][a] = true;
	}
	int E = n / 3;
	for (int i = 1; i <= n; i++){
		if (vis[i]) continue;
		for (int j = i + 1; j <= n; j++){
			if (vis[j] || G[i][j]) continue;
			//printf("(%d,%d)\n",i,j);
			vis[i] = vis[j] = true;
			break;
		}
	}
	int cnt = 0;
	REP(i,n){
		if (!vis[i]) printf("%d ",i),cnt++;
		if (cnt >= E) break;
	}
	return 0;
}