CF140CNew Year Snowmen

CF140C

题目大意:堆雪人,需要三个大小不同的雪球,现有(n)个给定大小的雪球，问最多堆多少个雪人

一个很明显的思路是把每种雪球出现的个数记录下来,然后直接扔到大根堆里面,每次选择剩下出现次数最多的三个堆成一个雪人,可以证明,这样一定不会比选择小的更劣

#include<cstdio>
#include<iostream>
#include<queue>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<vector>
#include<ctime>
#define LL long long
#define pii pair<int,int>
#define mk make_pair
#define fi first
#define se second
using namespace std;
const int N = 2e5 + 3;
int a[N];
int n;
int s[5];
priority_queue <pii> q; 
struct node{
	int x,y,z;	
}qq[N];
inline int read(){
	int v = 0,c = 1;char ch = getchar();
	while(!isdigit(ch)){
		if(ch == '-') c = -1;
		ch = getchar();
	}
	while(isdigit(ch)){
		v = v * 10 + ch - 48;
		ch = getchar();
	}
	return v * c;
}
int main(){
	n = read();
	for(int i = 1;i <= n;++i) a[i] = read();
	sort(a + 1,a + n + 1);
	int res = 1;
	for(int i = 1;i <= n;++i){
		if(a[i] == a[i + 1]) res++;
		else{
			q.push(mk(res,a[i]));
			res = 1;
		}
	}int ans = 0;
	while(q.size() >= 3){
		pii k1 = q.top();q.pop();
		pii k2 = q.top();q.pop();
		pii k3 = q.top();q.pop();
		ans++;
		s[1] = k1.se,s[2] = k2.se,s[3] = k3.se;
		sort(s + 1,s + 4);
		qq[ans] = (node){s[3],s[2],s[1]};
		k1.fi--,k2.fi--,k3.fi--;
		if(k1.fi) q.push(k1);
		if(k2.fi) q.push(k2);
		if(k3.fi) q.push(k3);
	}
	printf("%d
",ans);
	for(int i = 1;i <= ans;++i) printf("%d %d %d
",qq[i].x,qq[i].y,qq[i].z);
    return 0;
}