poj3186 Treats for the Cows

Treats for the Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4041		Accepted: 2063

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

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.

题意：

给一个序列，只能从俩端出，出一个乘以天数。

分析：

本来想，一个序列，只能从俩端出，就只要比较俩端的大小就可以了，其实不然，看下面的测试数据：

101 1 102 100

按照我想的结果是713；

可是如果你首先找出102为第5天，接下来就是递推的过程了。

所以明显大很多811.

所以状态转移方程是 dp[i][j]=max(dp[i+1][j]+v[i]*(n-j+i),dp[i][j-1]+v[j]*(n-j+i));
n-j+i是天数，通过一个一个找出的规律。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int dp[2005][2005];
int main()
{
   int n,i,g,k,v[2005],j;
   while(~scanf("%d",&n))
   {
       memset(dp,0,sizeof(dp));
       for(i=0;i<n;i++)
         {
             scanf("%d",&v[i]);
               dp[i][i]=v[i]*n;//初始化，从最后出对往前推，假设每个都是最后出对。
         }
        for(k=1;k<=n;k++)
       {
           for(i=0;i<n-k;i++)
             {
                 j=i+k;
                 dp[i][j]=max(dp[i+1][j]+v[i]*(n-j+i),dp[i][j-1]+v[j]*(n-j+i));
                 //这里是从最后出队的开始往前推，因为只有最后出队的，
                 //i+1才会等于j。  
            }
       }
        printf("%d
",dp[0][n-1]);
   }
   return 0;
}
/*#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
    int i,t,a[2005],j,ans,n;
    while(~scanf("%d",&t))
    {
        ans=0;n=1;
      for(i=0;i<t;i++)
        scanf("%d",&a[i]);
        i=0;j=t-1;n=1;
        while(n<=t)
        {

          if(a[i]<a[j])
           {
                ans+=a[i]*n;
                printf("ans=%d
",ans);
                  i++;
           }
          else
             {
                 ans+=a[j]*n;
                  printf("ans=%d
",ans);
                 j--;
             }
           n++;
        }
        printf("%d
",ans);
    }
    return 0;
}*/