每日三题-Day5-B（POJ 3186 Treats for the Cows 区间DP）

Treats for the Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5874		Accepted: 3049

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

题目大意：

给你n个数，可以把这个数列，当成是一个双向队列，可以从队头和队尾出队。

如果这个数是第i个出队的，那么这个数乘以i，求所有数出队之后，得到的最大的总和。

思路：

区间DP

num[i]代表数列。

dp[i][j]代表第i~第j个数组成的队列完全出队的最大和。

那么，dp[i][j+1],在增加了一个数字之后，可能有以下两种情况：

1. 先出i，然后剩下的数，就是i+1~j+1了。那么这个数列最大和的出队方案，和dp[i+1][j+1]的出队方案是一样的，只不过，所有数都延后一次出队，因为第一个让i出队了。

所以，最终的总和应该是num[i]+dp[i+1][j+1]+num[i+1]+...+num[j+1]

2. 先出j+1，然后剩下的是i~j。自然，剩下的数也和dp[i][j]的出队方案一样。

总和为num[j+1]+dp[i][j]+num[i]+...+num[j]

不妨用一个sum[i]数组，记录num[0~i]的和。这样能用前缀和来有效降低时间复杂度。

所以，状态转移方程为：

dp[i][j]=max(dp[i+1][j],dp[i][j-1])+sum[j]-sum[i-1];

AC代码：

#include
#include
using namespace std;

int dp[2002][2002];
int num[2002];
int sum[2002];

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i

伏见老贼又开新坑，祝贺《工口漫画老师》继漫画化之后又动画化并取得佳绩~（还是觉得妖精好可怜，求给个好结局吧）