cf442C Artem and Array

最新推荐文章于 2020-10-19 21:11:49 发布

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最新推荐文章于 2020-10-19 21:11:49 发布

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原文链接：http://www.cnblogs.com/zhber/p/4035960.html

本文介绍了一个CF比赛题目，通过使用贪心策略解决。问题涉及到一个整数数组，目标是在遵循特定规则的情况下获得最大分数。文章详细解释了解决方案，并提供了一个C++代码示例。

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Artem has an array of n positive integers. Artem decided to play with it. The game consists of n moves. Each move goes like this. Artem chooses some element of the array and removes it. For that, he gets min(a, b) points, where a and b are numbers that were adjacent with the removed number. If the number doesn't have an adjacent number to the left or right, Artem doesn't get any points.

After the element is removed, the two parts of the array glue together resulting in the new array that Artem continues playing with. Borya wondered what maximum total number of points Artem can get as he plays this game.

Input

The first line contains a single integer n (1 ≤ n ≤ 5·105) — the number of elements in the array. The next line contains n integers ai (1 ≤ ai ≤ 106) — the values of the array elements.

Output

In a single line print a single integer — the maximum number of points Artem can get.

Sample test(s)

input
5
3 1 5 2 6

output
11

input
5
1 2 3 4 5

output
6

input
5
1 100 101 100 1

output
102

贪心啊……

爆出内幕：昨天cf比赛的时候，zld大神认为此题贪心，结果没开long long第10个点wa。于是卓神信誓旦旦的说，这题贪心有反例。呵呵呵……

首先“填坑”：把两边都比中间大的数直接合掉，结果它就变成单调栈……

维护单调栈，最后sort一遍更新答案

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,top,zhan[500001];
long long ans;
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
	{
		int x;
		scanf("%d",&x);
		while (top>1 && zhan[top-1]>=zhan[top]&& zhan[top]<=x) 
		{
			ans+=min(zhan[top-1],x);
			top--;
		}
		zhan[++top]=x;
	}
	sort(zhan+1,zhan+top+1);
	for (int i=1;i<=top-2;i++)
	{
		ans+=zhan[i];
	}
	printf("%lld",ans);
}

转载于:https://www.cnblogs.com/zhber/p/4035960.html