程式師世界 >> 編程語言 >> C語言 >> C++ >> 關於C++ >> poj3666(Making the Grade)

poj3666(Making the Grade)

編輯：關於C++

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

Sample Output

題意：給定一個正整數序列a[1...n]，要求你改變每一個數變成c[1...n]，使得改變後的序列非嚴格單調，改變的代價為sigma(abs(a[i]-c[i]))，求代價最小值。

題解：顯然c[i]必定為a[1...n]中的某個值，且由於a過大，所以離散化，將a數組保存在b數組中並排序。設dp[i][j]為第i個數改變為b[j]時代價最小值，則dp[i][j]=min(dp[i-1][k])+abs(a[i]-b[j])，其中k<=j，即b[k]

#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

int n, a[2010], b[2010], dp[2010];

int main()
{
    cin >> n;
    for (int i = 0; i < n; ++i)
    {
        scanf(%d, &a[i]);
        b[i] = a[i];
    }
    sort(b, b+n);
    for (int i = 0; i < n; ++i)
        dp[i] = abs(a[0]-b[i]);
    for (int i = 1; i < n; ++i)
    {
        int tmp_min = dp[0];
        for (int j = 0; j < n; ++j)
        {
            tmp_min = min(tmp_min, dp[j]);
            dp[j] = tmp_min + abs(a[i]-b[j]);
        }
    }
    cout << *min_element(dp, dp+n) << endl;
    return 0;
}