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傳送門：

http://poj.org/problem?id=3186

Treats for the Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7672		Accepted: 4059

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 分析： 題意：給出的一系列的數字，可以看成一個雙向隊列，每次只能從隊首或者隊尾出隊，第n個出隊就拿這個數乘以n，最後將和加起來，求最大和 思路：由裏向外逆推區間 dp[i][j]表示左邊取了i個數，右邊取了j個數
故dp[i][j] = max(dp[i-1][j] + a[i]* (i+j), dp[i][j-1] + a[n-j+1]*(i+j));
註意當ij為0的邊界判斷即可。
code：

#include<stdio.h>
#include<string.h>
#include<memory>
using namespace std;
#define max_v 2005
int a[max_v];
int dp[max_v][max_v];
int main()
{
    /*
    dp[i][j]表示左邊取了i個數，右邊取了j個數
    故
    dp[i][j] = max(dp[i-1][j] + a[i]* (i+j), dp[i][j-1] + a[n-j+1]*(i+j));
    註意當ij為0的邊界判斷即可。
    */
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        memset(dp,0,sizeof(dp));
        int ans=0;
        for(int i=0;i<=n;i++)
        {
            for(int j=0;j+i<=n;j++)
            {
                if(i==0&&j==0)
                    dp[i][j]=0;
                else if(i==0)
                    dp[i][j]=dp[i][j-1]+a[n-j+1]*j;
                else  if(j==0)
                    dp[i][j]=dp[i-1][j]+a[i]*i;
                else
                    dp[i][j]=max(dp[i-1][j]+a[i]*(i+j),dp[i][j-1]+a[n-j+1]*(i+j));
            }
           ans=max(ans,dp[i][n-i]);
        }
        printf("%d\n",ans);
    }
    return 0;
}

POJ 3186 Treats for the Cows（區間dp）