POJ3666-Making the Grade

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本文介绍了一种解决修路问题的算法，旨在通过调整道路的海拔高度使其形成连续的单调斜坡，从而减少牲畜攀爬过程中的起伏。算法采用离散化处理和动态规划方法，计算出最小的修路成本。

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Making the Grade

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7185		Accepted: 3325

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

| A ₁- B ₁| + | A ₂- B ₂| + ... + | 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

Sample Output

Source

USACO 2008 February Gold

题意：有人想修一条尽量平缓的路，要么单调递增，要么单调递减，路的每一段海拔是A[i]，若修理后是B[i]，花费 | A[i] – B[i] | ，求最小花费

解题思路：最后不管怎么改变,最终的所有数都是原来的某个数，因为A[i]较大，所以可以对a[i]进行离散化，dp[i][j]表示第i个数变为原序列第j大的最小费用

#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <string>  
#include <algorithm>  
#include <cmath>  
#include <map>  
#include <cmath>  
#include <set>  
#include <stack>  
#include <queue>  
#include <vector>  
#include <bitset>  
#include <functional>  

using namespace std;

#define LL long long  
const LL INF = 0x3f3f3f3f3f3f3f3f;

int n, m;
int a[2005], b[2005];
LL dp[2005][2005];

int main()
{
	while (~scanf("%d", &n))
	{
		for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
		sort(b + 1, b + n + 1);
		int m = unique(b + 1, b + 1 + n) - b;
		for (int i = 1; i <= m - 1; i++) dp[1][i] = abs(b[i] - a[1]);
		for (int i = 2; i <= n; i++)
		{
			LL temp = INF;
			for (int j = 1; j <= m - 1; j++)
			{
				temp = min(temp, dp[i - 1][j]);
				dp[i][j] = abs(b[j] - a[i]);
				dp[i][j] += temp;
			}
		}
		LL ans = INF;
		for (int i = 1; i <= m - 1; i++) ans = min(ans, dp[n][i]);
		for (int i = 1; i <= m - 1; i++) dp[1][i] = abs(b[i] - a[1]);
		for (int i = 2; i <= n; i++)
		{
			LL temp = INF;
			for (int j = m-1; j >= 1; j--)
			{
				temp = min(temp, dp[i - 1][j]);
				dp[i][j] = abs(b[j] - a[i]);
				dp[i][j] += temp;
			}
		}
		for (int i = 1; i <= m - 1; i++) ans = min(ans, dp[n][i]);
		printf("%lld\n", ans);
	}
	return 0;
}