POJ-3666-Making the Grade(DP)

链接:

https://vjudge.net/problem/POJ-3666

题意:

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, ... , AN (1 ≤ N ≤ 2,000) describing the elevation (0 ≤ Ai ≤ 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, . ... , BN 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| + ... + | AN - BN |
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.

思路:

Dp[i, j], 表示第i个值, 变成j的最小花费,
考虑总和较大, 离散化把值控制到2000以内即可.

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#include <assert.h>
#include <iomanip>
#include <iostream>
#include <sstream>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;
const LL MOD = 20090717;
const LL MAXN = 2e3+10;

LL a[MAXN], Dp[MAXN][MAXN];
LL b[MAXN];
int n;

LL Abs(LL l, LL r)
{
    return l > r?l-r:r-l;
}

int main()
{
    scanf("%d", &n);
    for (int i = 1;i <= n;i++)
        scanf("%lld", &a[i]), b[i] = a[i];
    sort(b+1, b+1+n);
    for (int i = 1;i <= n;i++)
    {
        LL mmin = Dp[i-1][1];
        for (int j = 1;j <= n;j++)
        {
            mmin = min(mmin, Dp[i-1][j]);
            int val = Abs(b[j], a[i])+mmin;
            Dp[i][j] = val;
        }
    }
    LL res = Dp[n][1];
    for (int i = 1;i <= n;i++)
        res = min(res, Dp[n][i]);
    printf("%d
", res);

    return 0;
}