Treats for the Cows
FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for giving vast amounts of milk. FJ sells one treat per day and wants to maximize the money he receives over a given period time.
The treats are interesting for many reasons:
The first treat is sold on day 1 and has age a=1. Each subsequent day increases the age by 1.
The treats are interesting for many reasons:
- The treats are numbered 1..N and stored sequentially in single file in a long box that is open at both ends. On any day, FJ can retrieve one treat from either end of his stash of treats.
- Like fine wines and delicious cheeses, the treats improve with age and command greater prices.
- The treats are not uniform: some are better and have higher intrinsic value. Treat i has value v(i) (1 <= v(i) <= 1000).
- Cows pay more for treats that have aged longer: a cow will pay v(i)*a for a treat of age a.
The first treat is sold on day 1 and has age a=1. Each subsequent day increases the age by 1.
Input
Line 1: A single integer, N
Lines 2..N+1: Line i+1 contains the value of treat v(i)
Lines 2..N+1: Line i+1 contains the value of treat v(i)
Output
Line 1: The maximum revenue FJ can achieve by selling the treats
Sample Input
5
1
3
1
5
2
Sample Output
43
Hint
Explanation of the sample:
Five treats. On the first day FJ can sell either treat #1 (value 1) or treat #5 (value 2).
FJ sells the treats (values 1, 3, 1, 5, 2) in the following order of indices: 1, 5, 2, 3, 4, making 1x1 + 2x2 + 3x3 + 4x1 + 5x5 = 43.
Five treats. On the first day FJ can sell either treat #1 (value 1) or treat #5 (value 2).
FJ sells the treats (values 1, 3, 1, 5, 2) in the following order of indices: 1, 5, 2, 3, 4, making 1x1 + 2x2 + 3x3 + 4x1 + 5x5 = 43.
题意:
给一个数组v,每次可以取前面的或者后面的,第k次取的v[i]价值为v[i]*k,问总价值最大是多少。
题解:
区间DP
设dp[i][j]为从i取到j的最大值,假设每个数都是最后一个被取,那么dp[i][i]=a[i]*n。而dp[i][j]只能由dp[i+1][j]和dp[i][j-1]转移来,所以状态转移方程为dp[l][r] = Math.max(dp[l][r], Math.max(dp[l+1][r]+a[l]*(n-len), dp[l][r-1]+a[r]*(n-len)));其中(n-len)表示第几个取。
1 import java.util.Scanner;
2
3 public class Main {
4 static int casen,n;
5 static int [][] dp ;
6 static int [] a = new int [2019];
7 public static void main(String[] args) {
8 Scanner cin = new Scanner(System.in);
9 n = cin.nextInt();
10 dp = new int [n+10][n+10];
11 for(int i=1;i<=n;i++) {
12 a[i] = cin.nextInt();
13 dp[i][i] = a[i]*n;
14 }
15 long ans = 0;
16 for(int len=1;len<=n;len++) {
17 for(int l=1;l+len<=n;l++) {
18 int r = l+len;
19 dp[l][r] = Math.max(dp[l][r], Math.max(dp[l+1][r]+a[l]*(n-len), dp[l][r-1]+a[r]*(n-len)));
20 }
21 }
22 System.out.println(dp[1][n]);
23 }
24 }
25