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 a__i(1 ≤ a__i ≤ 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
并不会做.
接触了一个交单调栈的东西…
又是单调栈又是单调队列
是时候仔细了解下了.
1
2
3 /*************************************************************************
4 > File Name: code/2015summer/#3/D.cpp
5 > Author: 111qqz
6 > Email: rkz2013@126.com
7 > Created Time: 2015年07月28日 星期二 13时47分48秒
8 ************************************************************************/
9
10 #include<iostream>
11 #include<iomanip>
12 #include<cstdio>
13 #include<algorithm>
14 #include<cmath>
15 #include<cstring>
16 #include<string>
17 #include<map>
18 #include<set>
19 #include<queue>
20 #include<vector>
21 #include<stack>
22 #define y0 abc111qqz
23 #define y1 hust111qqz
24 #define yn hez111qqz
25 #define j1 cute111qqz
26 #define tm crazy111qqz
27 #define lr dying111qqz
28 using namespace std;
29 #define REP(i, n) for (int i=0;i<int(n);++i)
30 typedef long long LL;
31 typedef unsigned long long ULL;
32 const int N=5E5+7;
33 int a[N],b[N];
34 int n;
35 int main()
36 {
37 cin>>n;
38 int x;
39 int top = -1;
40 LL ans = 0;
41 for ( int i = 1 ; i <= n ; i++ )
42 {
43 scanf("%d",&x);
44 while (top>=1&&a[top-1]>=a[top]&&a[top]<=x)
45 {
46 ans = ans + min(x,a[top-1]);
47 top--;
48 }
49 top++;
50 a[top]=x;
51 }
52
53
54 sort(a,a+top+1);
55 for ( int i = 0 ; i <top-1 ; i++)
56 {
57 ans = ans + a[i];
58 }
59 cout<<ans<<endl;
60
61
62 return 0;
63 }