Kirill the Gardener 3 URAL - 2072

2072. Kirill the Gardener 3

Time limit: 2.0 second
Memory limit: 64 MB

Kirill the gardener has got a new task. He has to water the flowers growing on the huge flowerbed! It should be mentioned that the bed is very long and narrow at the same time. So from bird’s-eye view (and Kirill growth too) it looks like a straight line, where n points-flowers are located at regular intervals. For this job Kirill has a watering pot of infinite volume and smartwatch that shows moisture content of each flower before watering. The watering takes too much time, so the most dried flowers can die, which is unacceptable. So Kirill decided to water the flowers in order of their non-decreasing dryness. On the other hand, he wants to finish the watering as soon as possible, because there are a lot of other interesting things.

Assume that watering of one flower and walking between two neighbor flowers takes Kirill one minute. Can you figure out the time in which the young gardener will complete his job if he acts optimally? Initially Kirill stands near the leftmost flower.

Input

The first line contains an integer  n (1 ≤  n ≤ 10 5) — it’s amount of flowers in the flowerbed. The second line contains  n integers separated by spaces — it‘s moisture content of flowers given in order of their positions in the flowerbed from left to right. Moisture content is an integer from 1 up to 10 9 (including both).

Output

In the only line print an integer — the minimal time in which Kirill would complete watering.

Sample

input	output
6 3 2 5 6 2 5	21

Notes

There is one of the possible ways to finish up in 21 minutes in the example:

1. Go from the 1st to the 5th flower (4 minutes)
2. Water the 5th flower (1 minute)
3. Go from the 5th to the 2nd flower (3 minutes)
4. Water the 2nd flower (1 minute)
5. Go from the 2nd to the 1st flower (1 minute)
6. Water the 1st flower (1 minute)
7. Go from the 1st flower to the 3rd flower (2 minutes)
8. Water the 3rd flower (1 minute)
9. Go from the 3rd flower to the 6th flower (3 minutes)
10. Water the 6th flower (1 minute)
11. Go from the 6th flower to the 4th flower (2 minutes)
12. Water the 4th flower (1 minute)

Problem Author: Ilya Kuchumov
Problem Source: Ural Regional School Programming Contest 2015

Tags:  dynamic programming   ( hide tags for unsolved problems)

Difficulty: 183     Printable version     Submit solution     Discussion (5)

All submissions (2728)    All accepted submissions (905)    Solutions rating (567)

#include<bits/stdc++.h>
#include<algorithm>
#define ll long long
using namespace std;
const int maxn = 1e5+7;
int n, tot;
ll dp[maxn][2];
/*
   （从小到大）按照权值排序，再按照位置排序;
   分成若干区间d[i]：保存区间左右端点;(中间的一定会经过);
   dp[i][0]:表示从区间左端点到右端点;
   dp[i][1]:表示从区间右端点到左端点;
   状态转移方程：
   dp[i][0] = min(dp[i - 1][0] + abs(d[i - 1].rx - d[i].lx), dp[i - 1][1] + abs(d[i - 1].lx - d[i].lx)) + sum;
   dp[i][1] = min(dp[i - 1][0] + abs(d[i - 1].rx - d[i].rx), dp[i - 1][1] + abs(d[i - 1].lx - d[i].rx)) + sum;
   sum = abs(d[i].rx - d[i].lx);
*/

struct node{
  int lx, rx;
}id[maxn], d[maxn];

bool cmp(struct node a, struct node b)
{
    if(a.lx != b.lx) return a.lx < b.lx;
    else return a.rx < b.rx;
}


int main()
{
    //freopen("in.txt", "r", stdin);
    while(~scanf("%d", &n))
    {
        tot = 1;
        for(int i = 0; i < n ; i++) {
            id[i].rx = i;
            scanf("%d", &id[i].lx);
        }
        sort(id, id+n, cmp);
        d[tot].lx = d[tot].rx = id[0].rx;
        for(int i = 1; i < n; i++)
        {
            if(id[i].lx == id[i - 1].lx) d[tot].rx = id[i].rx;
            else {
                tot++;
                d[tot].lx = d[tot].rx = id[i].rx;
            }
        }
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= tot; i++)
        {
            int sum = abs(d[i].rx - d[i].lx);
            dp[i][0] = min(dp[i - 1][0] + abs(d[i - 1].rx - d[i].lx), dp[i - 1][1] + abs(d[i - 1].lx - d[i].lx)) + sum;
            dp[i][1] = min(dp[i - 1][0] + abs(d[i - 1].rx - d[i].rx), dp[i - 1][1] + abs(d[i - 1].lx - d[i].rx)) + sum;
        }
        printf("%lld\n",min(dp[tot][0], dp[tot][1]) + n);
    }
    return 0;
}