poj 3186 Treats for the Cows（区间dp）

Description

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 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.

Given the values v(i) of each of the treats lined up in order of the index i in their box, what is the greatest value FJ can receive for them if he orders their sale optimally? 

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)

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.

Source

USACO 2006 February Gold & Silver

 

题意：大概意思是，一批红酒开始有自身的价值，每天卖出第一个或最后一个，并且卖出的价值等于 原先的价值*存在的年数，求卖完所有的红酒能获得的最大价值

 

看完下面的提示立刻就想到了区间dp，dp[i][j]表示从i到j的最大价值，状态转移方程为dp[i][j]=max（dp[i+1][j]+a[i]*year,dp[i][j-1]+a[j]*a[j]*year）；

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<set>
#include<algorithm>
#include<cmath>
#include<stdlib.h>
#include<map>
using namespace std;
#define N 2006
int n;
int a[N];
int dp[N][N];
int main()
{
    while(scanf("%d",&n)==1)
    {
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++) 
        {
           scanf("%d",&a[i]); 
           dp[i][i]=a[i]*n;
          }
    
        for(int len=1;len<n;len++)
        {
            for(int i=1;i+len<=n;i++)
            {
                int j=i+len;
                dp[i][j]=max(dp[i+1][j]+a[i]*(n-(j-i+1)+1),dp[i][j-1]+a[j]*(n-(j-i+1)+1));
            }
        }
        
        
        printf("%d
",dp[1][n]);
    }
    return 0;
}

另外一种写法

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<set>
#include<algorithm>
#include<cmath>
#include<stdlib.h>
#include<map>
using namespace std;
#define N 2006
int n;
int a[N];
int dp[N][N];
int main()
{
    while(scanf("%d",&n)==1)
    {
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        
        memset(dp,0,sizeof(dp));
        for(int i=n;i>=1;i--)
        {
            for(int j=i;j<=n;j++)
            {
                   dp[i][j]=max(dp[i+1][j]+a[i]*(n-(j-i+1)+1),dp[i][j-1]+a[j]*(n-(j-i+1)+1));
                }
        }
        printf("%d
",dp[1][n]);
    }
    return 0;
}