POJ 3186 Treats for the Cows 简单DP

题目链接：http://poj.org/problem?id=3186

题目大意：给你N个数字，排列方式已经确定。现在你取数字，每次只能从头部或者尾部取一个数字，第k次取那么这个数字就乘以k，最后所得价值为N次总和。问能得到的最大值是多少？

解题思路：乍一看似乎是贪心，但是考虑 11 2 30 10，如果贪心的话是158，但是实际上，答案应当是11 2 10 30，即165. 那么如何DP呢？

决策：当前序列选择头部或者尾部。但这个决策实际上是个“我为人人”型，所以我们反过来，考虑序列为从第i个数到第j个数，那么其结果应该是从长度-1的序列递推得到，即i + 1到j或者i到j-1，而需要增加的值也可以根据其长度推出来，那么显然有：

dp[i][j]:=从第i个数到第j个数长度为j-i+1的序列可以得到的最大值，则

dp[i][i] = v[i] * n　　1 <= i <= j

dp[i][j] = max(dp[i + 1][j] + v[i] * age, dp[i][j - 1] + v[j] * age);　　age = (n + i - j)

代码：

const int inf = 0x3f3f3f3f;
const int maxn = 2e3 + 5;
int a[maxn], n;
int dp[maxn][maxn];

void solve(){
    memset(dp, 0, sizeof(dp));
    for(int i = 1; i <= n; i++) dp[i][i] = a[i] * n;
    for(int i = n - 1; i >= 0; i--){
        for(int j = i + 1; 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));
        }
    }
    printf("%d\n", dp[1][n]);
}
int main(){
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    solve();
}

题目：

Treats for the Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6347		Accepted: 3327

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