CodeForces 140C - New Year Snowmen

最新推荐文章于 2020-06-04 22:18:34 发布

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本文探讨如何通过合理选择不同大小的雪球来最大化制作雪人的数量，涉及从输入的雪球半径集合中找出三个不同半径的雪球进行组合。通过去重和排序等步骤，实现对输入数据的有效利用，最终输出能够构建的最大雪人数目及其具体组成。此过程展示了在有限资源条件下进行有效组合优化的方法。

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As meticulous Gerald sets the table and caring Alexander sends thepostcards, Sergey makes snowmen. Each showman should consist of threesnowballs: a big one, a medium one and a small one. Sergey's twins help him:they've already made n snowballs with radii equal to r₁, r₂, ..., r_n. To make a snowman, one needs any three snowballs whose radii arepairwise different. For example, the balls with radii 1, 2 and 3 can be used to make a snowman but 2, 2, 3 or 2, 2, 2 cannot. Help Sergey and his twins to determinewhat maximum number of snowmen they can make from those snowballs.

Input

The first line contains integer n (1 ≤ n ≤ 10⁵) — the number of snowballs. Thenext line contains n integers — the balls' radii r₁,r₂, ..., r_n (1 ≤ r_i ≤ 10⁹). The balls' radii can coincide.

Output

Print on the first line a single number k — the maximum number of the snowmen. Next k lines should contain the snowmen's descriptions.The description of each snowman should consist of three space-separated numbers— the big ball's radius, the medium ball's radius andthe small ball's radius. It is allowed to print the snowmen in any order. Ifthere are several solutions, print any of them.

Sample test(s)

input

7
1 2 3 4 5 6 7

output

2
3 2 1
6 5 4

input

3
2 2 3

output

思路：

去重并记录个数，用优先队列或者集合都可以

不知道为什么用优先队列做了好久都WA，理论上是没问题的，后来上网搜搜不到用优先队列的做法的，只好参考别人用集合的代码了，不过学到了迭代器^_^

这道题是熟悉STL容器的好题

参考程序：https://snipt.net/KayvanMazaheri/cf-140c-new-year-snowmen/
程序：
#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
#include <queue>
#include <map>
#include <set>
using namespace std;

#define L(u) (u<<1)
#define R(u) (u<<1|1)
#define lowbit(x) (x&-x)
#define rep(i,x,y) for (i=x;i<=y;i++)
#define ll __int64
#define max(x,y) ((x>y)?(x),(y))
#define min(x,y) ((x<y)?(x),(y))
#define sd(x) scanf("%d",&x)
#define sd2(x,y) scanf("%d%d",&x,&y)
#define slld(x) scanf("%lld",&x)

const int N = 100005;

struct node
{
	int x;
	int num;
	friend bool operator <(node a, node b)
	{
		return a.x<b.x;
	}
};

map<int, int> m;
struct CMP
{
	bool operator()(const int &a, const int &b)const
	{
		if (m[a] == m[b])
			return a>b;
		else
			return m[a]>m[b];
	}
};

typedef set<int, CMP>::iterator ITR;
set<int, CMP> q;
vector<int> ans;

int main()
{
	int n, i;
	sd(n);

	rep(i, 0, n - 1)
	{
		int temp;
		sd(temp);

		q.erase(temp);
		m[temp]++;
		q.insert(temp);
	}

	while (q.size()>2)
	{
		int a[3];
		rep(i, 0, 2)
		{
			ITR it = q.begin();
			a[i] = *it;

			q.erase(it);
			m[*it]--;
		}

		rep(i, 0, 2)
			if (m[a[i]]>0)
				q.insert(a[i]);

		sort(a, a + 3);

		ans.push_back(a[2]);
		ans.push_back(a[1]);
		ans.push_back(a[0]);
	}

	printf("%d\n", ans.size() / 3);
	for (i = 0; i<ans.size(); i += 3)
		printf("%d %d %d\n", ans[i], ans[i + 1], ans[i + 2]);

	return 0;
}