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题目链接：http://poj.org/problem?id=3186
Treats for the Cows
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5731 | Accepted: 2964 |
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 first treat is sold on day 1 and has age a=1. Each subsequent day increases the age by 1.
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.
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)
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.
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
dp一直都不太会推，，不过这道题和最长公共子序列有点像（可以看一下最长公共子序列，那个题是要找相同的，所以当找到相同的字符时，两个下标同时往后移一位）
还得加紧练习动态规划！！！
AC代码：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;int a[2005];
int dp[2005][2005];int main()
{int n,i,j;scanf("%d",&n);for(i=1;i<=n;i++)scanf("%d",&a[i]);memset(dp,0,sizeof(dp));for(i=0;i<=n;i++){for(j=0;j<=n-i;j++){if(i==0&&j==0)//注意当都没有取时，初始化为0，一定要记得跳过这一步continue;if(i==0&&j!=0)dp[i][j]=max(dp[i][j],dp[i][j-1]+(i+j)*a[n-j+1]);else if(i!=0&&j==0)dp[i][j]=max(dp[i][j],dp[i-1][j]+(i+j)*a[i]);elsedp[i][j]=max(dp[i-1][j]+(i+j)*a[i],dp[i][j-1]+(i+j)*a[n-j+1]);}}int ans=0;for(i=1;i<=n;i++)ans=max(ans,dp[i][n-i]);//这儿不能确定是哪一个dp[][]是最大的，因为i,j遍历的是同一个序列printf("%d\n",ans);return 0;
}
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