cf442C Artem and Array

C. Artem and Array

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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);
}