(离散) Making the grade (P3666)_
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题意:给出一个序列,,单独的一个数可以任意加减,
求把这样的序列变成最长不减序列时对每个数的相加或相减的数目之和
思路:1、把给定的序列排好序,就成为其离散的不下降的点
2、对第一点进行与排好序的最小值的点进行比较,求得dp[0][j]要升到第j的高度时所要的花费
3、对第二点及以后的每一点进行更新。dp[i][j]第i+1点到高度j时的前i+1个总的花费
4、找到更后最后一个点到任意高度的最小值便为答案
#include<iostream>#include<algorithm>using namespace std;#define N 2003int n;int go[N],hi[N];int dp[N][N];bool cmp(int a,int b){return a>b;}int main(){freopen("in.txt","r",stdin);int i,j,k;cin>>n;for (i=0;i<n;i++)cin>>go[i],hi[i]=go[i];sort(hi,hi+n);for (i=0;i<n;i++)dp[0][i]=abs(hi[i]-go[1]);for (i=1;i<n;i++){k=dp[i-1][0];for (j=0;j<n;j++){k=min(k,dp[i-1][j]);dp[i][j]=k+abs(hi[j]-go[i]);}}int ans=dp[n-1][0];for (i=0;i<n;i++)ans=min(ans,dp[n-1][i]);cout<<ans<<endl;return 0;}
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 ofN 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 sequenceB1, . ... , 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.
Sample Input
71324539
Sample Output
3
Source
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