POJ 3666 Making the Grade （dp）

Making the Grade

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4404 Accepted: 2094

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

71324539

Sample Output

3

dp[i][j]表示第i个位置的值改成b数组中第j个数的值，且前i个有序（单调不上升或者单调不下降）
dp[i][j] = min (dp[i - 1][k]）+ abs (a[i] - b[j])     最小的dp[i - 1][k]可以边循环边找，且k<=j

然后分别做一次就可以了

AC代码如下：

////  POJ 3666 Making the Grade////  Created by TaoSama on 2015-03-06//  Copyright (c) 2015 TaoSama. All rights reserved.//#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>#define CLR(x,y) memset(x, y, sizeof(x))using namespace std;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int N = 1e5 + 10;//dp[i][j]表示第i个位置的值改成b数组中第j个数的值，且前i个有序（单调不上升或者单调不下降）//dp[i][j] = min (dp[i - 1][k]）+ abs (a[i] - b[j])//最小的dp[i - 1][k]可以边循环边找，且k<=jint n, a[2005], b[2005], dp[2005];int solve() {memset(dp, 0, sizeof dp);for(int i = 1; i <= n; ++i) {int preMin = INF;for(int j = 1; j <= n; ++j) {preMin = min(preMin, dp[j]);dp[j] = preMin + abs(a[i] - b[j]);}}return *min_element(dp + 1, dp + 1 + n);}int main() {#ifdef LOCALfreopen("in.txt", "r", stdin);//freopen("out.txt","w",stdout);#endifios_base::sync_with_stdio(0);while(cin >> n) {for(int i = 1; i <= n; ++i) cin >> a[i], b[i] = a[i];sort(b + 1, b + 1 + n);int ans = INF;ans = min(ans, solve());reverse(b + 1, b + 1 + n);ans = min(ans, solve());cout << ans << endl;}return 0;}