Treats for the Cows
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问题描述
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
- 5↵
- 1↵
- 3↵
- 1↵
- 5↵
- 2↵
- 43↵
题目等效含义:
给出一个数字序列,然后每次只可以从队首或是队尾取相应的数字并乘上次数,第几个取出,取出的元素就乘以几,然后将所有的乘完之后的元素加起来,求最大的和。
大致思路:
一开始的时候以为这道题就是一道贪心题目,因为对于样例贪心算法完全符合。但是交了之后才发现不对,例如这样一个用例8 1 9 7,如果按照贪心的思路是7*1+8*2+1*3+9*4=62,但是如果这样去9*4+7*3+1*2+8*1=67,显然贪心得到的结果是不对,所以就要利用动态规划的方法进行计算。我们可以利用由内到外的思路进行计算,在开始的时候假设每一个数字都可以是最后被取出的,然后依次向前利用动态规划进行计算,每一次比较前后乘以相应次数的结果的最大值,将其赋给当状态。最后输出dp[1][n]即可。
具体实现:
先将给出的数字存入一个数组里面,然后再定义一个二维数组表示相应的状态,用以记录取相应次数的时候得到的最大值。开始的时候假设每一个数字都可以是最后被取出的,然后利用一个两重循环,依此由内向外推即可。
注意事项:
(1)因为数字序列的长度比较大,所以定义的数组要当做全局变量,不然的话会re。
(2)状态转移方程注意数组的下标,注意数组下标的变换。
(3)注意最后的输出,根据自己的状态数组输出相应的一个值。
实现代码
<span style="font-family:Microsoft YaHei;font-size:14px;">#include<stdio.h>#include<string.h>int dp[2005][2005];//int max(int x,int y)//{//if(x>y)//return x;//return y;//}int main(){ int n,i,p,j; int jiazhi[2005]; int temp1,temp2; memset(dp,0,sizeof(dp)); scanf("%d",&n); for(i=0;i<n;i++) { scanf("%d",&jiazhi[i]); dp[i][i]=jiazhi[i]*n; } for(p=1;p<=n;p++) { for(i=0;i<n-p;i++) { j=i+p; temp1=dp[i+1][j]+jiazhi[i]*(n-j+i); temp2=dp[i][j-1]+jiazhi[j]*(n-j+i); // dp[i][j]=max(dp[i+1][j]+jiazhi[i]*(n-j+i),dp[i][j-1]+jiazhi[j]*(n-j+i)); if(temp1>temp2) { dp[i][j]=temp1; } else { dp[i][j]=temp2; } } } printf("%d\n",dp[0][n-1]); return 0;}</span>
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