http://poj.org/problem?id=3186
Treats for the Cows
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 9704 |
|
Accepted: 5021 |
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.
题意：给你n个数，每次从两端取，取完为止，取得每个数*取得顺序（1，2，... n），使该值最大为多少？
思路：区间dp，dp[i][j]表示从i到j区间取值最大，可以由dp[i+1][j] 和 dp[i][j-1] 这两区间转移而来
//#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
#define ME(x , y) memset(x , y , sizeof(x))
#define SF(n) scanf("%d" , &n)
#define rep(i , n) for(int i = 0 ; i < n ; i ++)
#define INF 0x3f3f3f3f
#define mod 998244353
#define PI acos(-1)
using namespace std;
typedef long long ll ;
int a[2009] , dp[2009][2009];
int main()
{
int n ;
scanf("%d" , &n);
for(int i = 1 ; i <= n ; i++)
{
scanf("%d" , &a[i]);
}
memset(dp , 0 , sizeof(dp));
for(int i = n ; i >= 1 ; i--)
{
for(int j = i ; j <= n ; j++)
{
dp[i][j] = max(dp[i+1][j] + a[i]*(n+i-j) , dp[i][j-1] + a[j]*(n+i-j));
}
}
cout << dp[1][n] << endl ;
return 0 ;
}
区间dp
原文：https://www.cnblogs.com/nonames/p/11779480.html
