POJ 3666 Making the Grade

Making the Grade

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5037		Accepted: 2384

Description

A straight dirt road connects two fields on FJ's farm, but it changes elevation more than FJ would like. His cows do not mind climbing up or down a single slope, but they are not fond of an alternating succession of hills and valleys. FJ would like to add and remove dirt from the road so that it becomes one monotonic slope (either sloping up or down).

You are given N integers A₁, ... , A_N (1 ≤ N ≤ 2,000) describing the elevation (0 ≤ A_i ≤ 1,000,000,000) at each of N equally-spaced positions along the road, starting at the first field and ending at the other. FJ would like to adjust these elevations to a new sequence B₁, . ... , B_N that is either nonincreasing or nondecreasing. Since it costs the same amount of money to add or remove dirt at any position along the road, the total cost of modifying the road is

|A₁ - B₁| + |A₂ - B₂| + ... + |A_N - B_N |

Please compute the minimum cost of grading his road so it becomes a continuous slope. FJ happily informs you that signed 32-bit integers can certainly be used to compute the answer.

Input

* Line 1: A single integer: N
* Lines 2..N+1: Line i+1 contains a single integer elevation: A_i

Output

* Line 1: A single integer that is the minimum cost for FJ to grade his dirt road so it becomes nonincreasing or nondecreasing in elevation.

Sample Input

7
1
3
2
4
5
3
9

Sample Output

3

这个题目苦思冥想了好久，但是一直不能总结出转移方程，看了这个人的题解。发现用DP求解这个真的很简单。

另外开一个数组保存排好序的输入的数字串

利用dp[i][j]表示前i个元素中以j作为最后一个数的单调序列需要的最小花费。

转移方程：dp[i][j]  = min(dp[i – 1][k]) + |A[i] – B[j]|       (k = 0…j)

其中如果我们每次对k都再进行一次循环会超时，我们发现k的值就是i - 1之前所有的数字以k为最后的最小消费。他和dp[i - 1][j]中的最小值就是k的值。就不用循环了。

代码如下：

/*************************************************************************
	> File Name: Making_the_Grade.cpp
	> Author: Zhanghaoran
	> Mail: chilumanxi@xiyoulinux.org
	> Created Time: Wed 28 Oct 2015 10:33:23 PM CST
 ************************************************************************/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#define ll long long
using namespace std;

int N;
ll a[2010];
ll dp[2010][2010];
ll temp[2010];

int min(ll a, ll b){
    return a < b ? a : b;
}

int main(void){
    scanf("%d", &N);
    for(int i = 1; i <= N; i ++){
        scanf("%lld", &a[i]);
        temp[i] = a[i];
    }
    sort(temp + 1, temp + 1 + N);
    for(int i  = 1; i <= N; i ++){
        dp[1][i] = llabs(a[1] - temp[i]);
        //cout << temp[i] << endl;
    }

    for(int i = 2; i <= N; i ++){
        ll k = dp[i - 1][1];
        for(int j = 1; j <= N; j ++){
            k = min(k, dp[i - 1][j]);
            dp[i][j] = k + llabs(a[i] - temp[j]);
        }
    }
    ll Min = dp[N][1];
    for(int i = 2; i <= N; i ++){
        if(dp[N][i] < Min)
            Min = dp[N][i];
    }
    
    printf("%d
", Min);
}