POJ 3186 Treats+for+the+Cows 简单dp或者区间dp

题目链接：POJ - 3186

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.

题意：一个序列可以从左边取也可以从右边取  第i次取的可以获得 i*v(i) 的贡献 (v(i)是第i次取得的权值) 求最大的贡献

第一种方法 由外向内推：

设 表示左边取完i个右边取完j个 的最大值 我们考虑一些如何转移 

边界条件    

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 2003;
typedef long long ll;
ll dp[N][N],a[N];
int main(){
	int n;
	scanf("%d",&n);
	for(int i = 1; i <= n; i++) scanf("%lld",&a[i]);
	dp[0][0]=0;
	for(int i = 1; i <= n; i++) 
	dp[i][0]=dp[i-1][0]+1ll*i*a[i],dp[0][i]=dp[0][i-1]+1ll*i*a[n-i+1];
	for(int i = 1; i <= n; i++){
		for(int j = 1; i+j <= n; j++)
		dp[i][j]=max(dp[i-1][j]+a[i]*1ll*(i+j),dp[i][j-1]+a[n-j+1]*1ll*(i+j));
	}
	ll ans = 0;
	for(int i = 0; i <= n; i++) 
	ans=max(dp[i][n-i],ans);
	printf("%lld\n",ans);
	return 0;
} 

第二种方法 由内向外推：

这个就是区间dp了 但是没有三重 只有两层 简单递推即可

表示区间 l 到 r 可以取到的最大值 

区间dp有一个精髓的地方在于是由小区间开始 去推大的区间 初始状态就是一个元区间（长度为1的区间）

我们考虑出状态的值是多少 显然 因为一个区间肯定是最后被取到的 所以 ：

考虑转移：

稍微有点丑 大家凑合着看  也就是一个区间要么是由 l+1 到 r 这个区间加上 a[ l ] 的贡献而来 要么是由 l 到 r-1 这个区间加上a[ r ] 而来 我们取最大值即可

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
int n;
ll dp[2003][2003],a[2003];
int main(){
	scanf("%d",&n);
	for(int i = 1; i <= n; i++) scanf("%lld",&a[i]);
	for(int i = 1; i <= n; i++) dp[i][i]=1ll*n*a[i];
	for(int i = 2; i <= n; i++){
		for(int l = 1; l <= n-i+1; l++){
			int r = l+i-1;
			dp[l][r]=max(dp[l+1][r]+1ll*(n-r+l)*a[l],dp[l][r-1]+1ll*(n-r+l)*a[r]);
		}
	}
	printf("%lld\n",dp[1][n]);
	return 0;
}