poj3666 Making the Grade

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, … , 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

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

Input

Line 1: A single integer: N

Lines 2..N+1: Line i+1 contains a single integer elevation: Ai

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.

首先说明贪心是错的。比如 2 3 4 5 6 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1…（很多个1）很明显如果要贪心需要把这些1都变成6，而正解显然是把所有数变成1。

其次是这个题数据比较弱，只需要算单调不递减就可以过【其实我本来不知道，AC了以后发现自己程序有问题，感到很奇怪，于是看了discuss】。

首先离散化，然后dp[i][j]表示前i个元素，最后一位不大于j的最小费用。

因为如果有一个元素可以接在大的后面，那一定更可以接在小的后面。

这样dp[i][j]=min(dp[i][j-1],dp[i-1][j]+cost)，cost即为把当前元素改为j的花费。

O(n^2)解决。

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
int a[2010],tem[2010],now[2010],dp[2010][2010];
int main()
{
int i,j,k,m,n,p,q,x,y,z;
scanf("%d",&n);
for (i=1;i<=n;i++)
scanf("%d",&a[i]),tem[i]=a[i];
sort(tem+1,tem+n+1);
m=unique(tem+1,tem+n+1)-tem-1;
for (i=1;i<=n;i++)
now[i]=lower_bound(tem+1,tem+m+1,a[n-i+1])-tem;
for (i=1;i<=n;i++)
dp[i][0]=0x3f3f3f3f;
for (i=1;i<=n;i++)
for (j=1;j<=m;j++)
dp[i][j]=min(dp[i][j-1],dp[i-1][j]+abs(tem[now[i]]-tem[j]));
printf("%d\n",dp
[m]);
}