poj3186 Treats for the Cows

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

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

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.

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：

4

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;
}*/