AOJ 224.Treats for the Cows

213

题目



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.


题解


题目是要不断从两段取数,乘上取得次序,最后所有数加起来和最大
按照样例,就是 前( 1 ) 后( 2 ) 前( 3 ) 前( 1 ) 后( 5 )
最后最大为 43

看上去似乎没思路,不过可以先看下关系
每一次从前或后取,压根不连续很难找到状态转移方程
为了连续可以倒着想这道题
从后往前找,也即向外拓展

dp[i][j] 表示 ij 的最大值
它来自于 dp[i+1][j]dp[i][j-1] 向外拓展
dp[i][j] = max{ dp[i + 1][j] + a[i] * (n - j + i) , dp[i][j - 1] + a[j] * (n - j + i) }

根据动态规划到达一个状态其之前的状态必须已经到达, i 应该逆序便利, j 正序遍历

代码


/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい?
エリーチカ!
要写出来Хорошо的代码哦~
*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <functional>
using namespace std;

const int maxn = 2005;

int a[maxn];
int dp[maxn][maxn];

bool Do() {
int n;
if(scanf("%d",&n) == EOF)
return false;
for(int i = 1;i <= n;i++)
scanf("%d",&a[i]);



for(int i = n;i >= 1;i--)
for(int j = i;j <= n;j++) {
if(i == j) {
dp[i][i] = a[i] * n;
} else {
dp[i][j] = max(
dp[i + 1][j] + a[i] * (n - j + i),
dp[i][j - 1] + a[j] * (n - j + i)
);
}
}

printf("%d\n",dp[1][n]);

return true;
}

int main() {
while(Do());
return 0;
}
发布评论
  • 点击查看/关闭被识别为广告的评论