poj3186——Treats for the Cows（区间dp）-白红宇

poj3186——Treats for the Cows（区间dp）

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发布时间：2019-05-10

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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

题意是给一串长度为n的数字序列，每次只能从队首或队尾出列，出列时对应的数字乘以出列顺序，最后全部加起来，求最大的值。

学习下区间dp。这种题一般都会枚举区间长度，然后向外按照题意扩展。

本体是从最后卖的物品开始，dp到最先卖出的物品，区间长度当然也是从小到大。

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             #define INF 0x3f3f3f3f#define MAXN 2005#define Mod 10001using namespace std;int a[MAXN],dp[MAXN][MAXN]; //dp[i][j]表示i~j这个区间的最优解int main(){ int n; while(~scanf("%d",&n)) { for(int i=1; i<=n; ++i) scanf("%d",&a[i]); memset(dp,0,sizeof(dp)); for(int i=1;i<=n;++i) dp[i][i]=n*a[i]; //初始条件，只有一个数的时候最优解是最后一天卖 for(int len=1; len

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