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.
题目大意：
给出的一系列的数字，可以看成一个双向队列，每次只能从队首或者队尾出队，第n个出队就拿这个数乘以n，最后将和加起来，求最大和
思路：从外向里推，并不是很好推， 于是应该从里向外逆推区间，这样就简单多了
记忆化搜索：
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
using namespace std;
typedef long long LL;
#define N 2005
#define met(a,b) (memset(a,b,sizeof(a))) int a[N], dp[N][N], n; int DFS(int L, int R, int k)
{
if(L>R || L< || R< || R>n || L>n) return -; if(dp[L][R]!=-)
return dp[L][R]; dp[L][R] = ;
dp[L][R] = max(DFS(L+, R, k+) + a[L]*k, DFS(L, R-, k+) + a[R]*k); return dp[L][R];
} int main()
{ while(scanf("%d", &n)!=EOF)
{
int i; met(dp, -);
met(a, ); for(i=; i<=n; i++)
scanf("%d", &a[i]); for(i=; i<=n; i++)
dp[i][i] = a[i]*n; dp[][n] = DFS(, n, ); printf("%d\n", dp[][n]);
}
return ;
}
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
using namespace std;
typedef long long LL;
#define N 2005
#define met(a,b) (memset(a,b,sizeof(a))) int a[N], dp[N][N], n; int main()
{ while(scanf("%d", &n)!=EOF)
{
int i, j, l; met(dp, );
met(a, ); for(i=; i<=n; i++)
scanf("%d", &a[i]); for(i=; i<=n; i++)
dp[i][i] = a[i]*n; for(l=; l<n; l++)
{
for(i=; i+l<=n; i++)
{
j = i+l;
dp[i][j] = max(dp[i+][j]+a[i]*(n-l), dp[i][j-]+a[j]*(n-l));
}
} printf("%d\n", dp[][n]);
}
return ;
}