poj3186(搜索)

Treats for the Cows

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5741 Accepted: 2968

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

513152

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

   与其说是搜索还不如说是暴力枚举....也看到有人用dp写，贴个代码。

#include<bits/stdc++.h>using namespace std;int ans[10000];int maxx[3005][3005];int t;int dfs(int st,int en){if(maxx[st][en]!=0) return maxx[st][en];if(st==en) return maxx[st][st]=t*ans[st];int x=dfs(st+1,en)+(t-en+st)*ans[st];int y=dfs(st,en-1)+(t-en+st)*ans[en];maxx[st][en]=max(x,y);return maxx[st][en];}int main(){    while(cin>>t){    memset(maxx,0,sizeof(maxx));    for(int i=1;i<=t;i++) scanf("%d",&ans[i]);cout<<dfs(1,t)<<endl;}}