【BZOJ】1652: [Usaco2006 Feb]Treats for the Cows（dp）

http://www.lydsy.com/JudgeOnline/problem.php?id=1652

dp。。

我们按间隔的时间分状态k，分别为1～n天

那么每对间隔为k的i和j。而我们假设i或者j在间隔时间内最后取。那么在这个间隔时间内最后取的时间就是n-k+1(这个自己想。。也就是说，之前在n-(k-1)+1的时间间隔内取过了，现在我们要多了一个时刻，相当于取这个早了一个时间）

然后就是

k为阶段

i为左端点

j=i+k-1为右端点

t=n-k+1为i-j取最后一个的时间

然后转移

f[i][j]=max(f[i+1][j]+t*a[i], f[i][j-1]+t*a[j])

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }

const int N=2005;
int f[N][N], n, a[N];

int main() {
	read(n);
	for1(i, 1, n) read(a[i]);
	for1(k, 1, n)
		for(int i=1; i+k-1<=n; ++i) {
			int t=n-k+1, j=i+k-1;
			f[i][j]=max(f[i+1][j]+t*a[i], f[i][j-1]+t*a[j]);
		}
	print(f[1][n]);
	return 0;
}

---

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.

约翰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．每天约翰售出一份零食．当然约翰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，约翰每

  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，这样约翰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．约翰进货时，第i份零食的初始价值为Vi(1≤Vi≤1000)．

  •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．

  Vi的是从盒子顶端往下的第i份零食的初始价值．约翰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．

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

Five treats. On the first day FJ can sell either treat #1 (value 1) or
treat #5 (value 2).

Sample Output

43

OUTPUT DETAILS:

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.

HINT

Source

Silver