区间dp

http://poj.org/problem?id=3186

Treats for the Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9704		Accepted: 5021

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.

题意：给你n个数，每次从两端取，取完为止，取得每个数*取得顺序（1，2，... n），使该值最大为多少？

思路：区间dp，dp[i][j]表示从i到j区间取值最大，可以由dp[i+1][j] 和 dp[i][j-1] 这两区间转移而来

//#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
#define ME(x , y) memset(x , y , sizeof(x))
#define SF(n) scanf("%d" , &n)
#define rep(i , n) for(int i = 0 ; i < n ; i ++)
#define INF  0x3f3f3f3f
#define mod 998244353
#define PI acos(-1)
using namespace std;
typedef long long ll ;
int a[2009] , dp[2009][2009];

int main()
{
    int n ;
    scanf("%d" , &n);
    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+i-j) , dp[i][j-1] + a[j]*(n+i-j));
        }
    }
    cout << dp[1][n] << endl ;
    return 0 ;
}