POJ-3186 Treats for the Cows( DP )

题目链接: http://poj.org/problem?id=3186

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.

 

题目并没有看明白，还是看最后的Hint才明白的(雾)，大意是，给一个数列，每次只能从头或从尾取一个数，第i次取的数乘i，求和最大为多少。

状态转移方程 以dp[i][j]表示列头取了i个数，列尾取了j个数，则dp[i][j]只与dp[i - 1][j]和dp[i][j - 1]有关，于是得到状态转移方程dp[i][j] = max( dp[i - 1][j] + val[i - 1] * ( i + j ), dp[i][j - 1] + val[n - j] * ( i + j ) ),注意边界情况，此处的val我是按下标从0开始的。

 

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

int val[2005];
int dp[2005][2005];

int main(){
    ios::sync_with_stdio( false );

    int n;
    while( cin >> n ){
        for( int i = 0; i < n; i++ )
            cin >> val[i];
        memset( dp, 0, sizeof( dp ) );

        dp[1][0] = val[0];
        for( int i = 2; i <= n; i++ )
            dp[i][0] = dp[i - 1][0] + val[i - 1] * i;

        dp[0][1] = val[n - 1];
        for( int i = 2; i <= n; i++ )
            dp[0][i] = dp[0][i - 1] + val[n - i] * i;

        for( int i = 1; i <= n; i++ )
            for( int j = 1; j + i <= n; j++ )
                dp[i][j] = max( dp[i - 1][j] + val[i - 1] * ( i + j ), dp[i][j - 1] + val[n - j] * ( i + j ) );

        int ans = 0;
        for( int i = 0; i <= n; i++ )
            ans = max( ans, dp[i][n - i] );

        cout << ans << endl;
    }

    return 0;
}