POJ 3186 Treats for the Cows(区间dp)

Treats for the Cows

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6180 Accepted: 3223

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.

场景就是类似于双端队列，你每次只能从两个端口取食物，求最大的价值

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <queue>using namespace std;typedef long long LL;const int inf = 0x3f3f3f3f;const LL INF = 0x3f3f3f3f3f3f3f3fll;const int maxn = 2e3+10;int N,a[maxn];int dp[maxn][maxn];int main(){while(~scanf("%d",&N)){for(int i = 1; i <= N; i++)scanf("%d",&a[i]);for(int day = 0; day <= N; day++){//现在剩余多少天没吃for(int i = 1; i+day <= N; i++){int j = i+day;//j-i == day,就是还有day天没花费dp[i][j] = max(dp[i+1][j]+a[i]*(N-day),dp[i][j-1]+a[j]*(N-day));//逆推}}printf("%d\n",dp[1][N]);}return 0;}