POJ 3186 Treats for the Cows

Treats for the Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7690		Accepted: 4074

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

Sample Output

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

USACO 2006 February Gold & Silver

算法分析：

题意：

在一个长度为n的区间里，你可以在两端（左端或者右端）取出一个数，这个数乘以他是第几次取出来的。求和的最大值。

分析：

dp[ i ][ j ]:表示在区间[ i , j ]（j>=i）里可以获得的最大值。

如果我们从里往外进行计算，找一个数以它作为最后取出来的，由题意可以知道区间[ i , j ]只可能由区间 [ i+1,j ]或者 [ i , j-1 ] 组成，结合题意

状态转移方程：

dp[i][j]=max(dp[i+1][j]+a[i]*(n+i-j),dp[i][j-1]+a[j]*(n+i-j));

代码实现：

#include<cstdio>  
#include<cstring>  
#include<cstdlib>  
#include<cctype>  
#include<cmath>  
#include<iostream>  
#include<sstream>  
#include<iterator>  
#include<algorithm>  
#include<string>  
#include<vector>  
#include<set>  
#include<map>  
#include<stack>  
#include<deque>  
#include<queue>  
#include<list>  
using namespace std;  
const double eps = 1e-8;  
typedef long long LL;  
typedef unsigned long long ULL;  
const int INT_INF = 0x3f3f3f3f;  
const int INT_M_INF = 0x7f7f7f7f;  
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;  
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;  
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};  
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1}; 
const long long BASE = 1e18; 
const int MOD = 1e9 + 7;  
const double pi = acos(-1.0);  
const int MAXN = 1e5 + 10;  
const int MAXT = 10000 + 10;  
const int M=2005;
using namespace std;
int dp[M][M],a[M];
int main()
{
	int n;
    while(scanf("%d",&n)!=EOF)
    {
    	memset(a,0,sizeof(a));
    	memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        	scanf("%d",&a[i]);
		for(int i=n;i>=1;i--) //这里i逆序的原因：dp[i][j]由dp[i+1][j]得来的，按照题意，它是有后端取出的
		 for(int j=i;j<=n;j++) //j正序原因同上
		 dp[i][j]=max(dp[i+1][j]+a[i]*(n+i-j),dp[i][j-1]+a[j]*(n+i-j));
		printf("%d\n",dp[1][n]);
    }
 
    return 0;
}

POJ 3186 Treats for the Cows

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