「CF442C」Artem and Array

2014年6月20日3,6702

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 getsmin(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

1 2	5 3 1 5 2 6

output

11

input

5
1 2 3 4 5

1 2	5 1 2 3 4 5

output

6

input

5
1 100 101 100 1

1 2	5 1 100 101 100 1

output

102

102

题解

首先填坑。。。也就是如果一个点俩边比它高把它删了。。。

然后序列整个序列递增或者递减。。。贪心即可

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#define inf 0x7fffffff
#define ll long long
using namespace std;
inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int s[500005];
int n,top;
ll ans;
int main()
{
	n=read();
	for(int i=1;i<=n;i++)
	{
		int x=read();
		while(top>1&&s[top-1]>=s[top]&&x>=s[top])
		    {ans+=min(s[top-1],x);top--;}
		s[++top]=x;
	}
	sort(s+1,s+top+1);
	for(int i=1;i<=top-2;i++)ans+=s[i];
	printf("%I64d",ans);
	return 0;
}