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

Sample Output

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\n",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\n",dp[1][n]);

     }

     return 0;

 }

巴特西

poj 3186 Treats for the Cows（区间dp）

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