SP740 TRT - Treats for the Cows
题目描述
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.
输入格式
Line 1: A single integer, N
Lines 2..N+1: Line i+1 contains the value of treat v(i)
输出格式
The maximum revenue FJ can achieve by selling the treats
题意翻译
题目描述
FJ经常给产奶量高的奶牛发特殊津贴，于是很快奶牛们拥有了大笔不知该怎么花的钱．为此，约翰购置了N(1≤N≤2000)份美味的零食来卖给奶牛们．每天FJ售出一份零食．当然FJ希望这些零食全部售出后能得到最大的收益．这些零食有以下这些有趣的特性：
•零食按照1．．N编号，它们被排成一列放在一个很长的盒子里．盒子的两端都有开口，FJ每天可以从盒子的任一端取出最外面的一个．
•与美酒与好吃的奶酪相似，这些零食储存得越久就越好吃．当然，这样FJ就可以把它们卖出更高的价钱．
•每份零食的初始价值不一定相同．FJ进货时，第i份零食的初始价值为Vi(1≤Vi≤1000)(Vi指的是从盒子顶端往下的第i份零食的初始价值)．
•第i份零食如果在被买进后的第a天出售，则它的售价是vi×a．
FJ告诉了你所有零食的初始价值，并希望你能帮他计算一下，在这些零食全被卖出后，他最多能得到多少钱．
输入输出格式
输入格式
第一行：一个整数n。 接下来的n行：每行一个整数Vi。
输出格式
FJ通过出售零食最多能得到的钱数。
输入输出样例
输入 #1
5
1
3
1
5
2
输出 #1
43
【思路】
区间DP
【题目大意】
每次都从左端点或者右端点选择一个零食
获得的价值是这个零食的价值乘以是第几个选择的
【核心思路】
可以设置一个状态f(i,j)
表示选取了i个零食，在左边选取了j个
这个状态可以由前面选取了i - 1个零食
在左边选取j个零食或者在左边选取了j-1个零食
意思就是：
之前选取了i-1个零食
现在选取的第i个零食
分别在左边选的还是在右边选的情况转移过来
【DP方程式】
\[f[i][j] = max(f[i - 1][j] + a[n - i + 1 + j] * i,f[i - 1][j - 1] + a[j] * i)\]
【完整代码】
#include<iostream>
#include<cstdio>
using namespace std;
const int Max = 2005;
int a[Max];
int f[Max][Max];
int main()
{
int n;
cin >> n;
for(register int i = 1;i <= n;++ i)
cin >> a[i];
for(register int i = 1;i <= n;++ i)
for(register int j = 0;j <= i;++ j)
f[i][j] = max(f[i - 1][j] + a[n - i + 1 + j] * i,f[i - 1][j - 1] + a[j] * i);
int M = 0;
for(register int i = 0;i <= n;++ i)
M = max(M,f[n][i]);
cout << M << endl;
return 0;
}