POJ3186:Treats for the Cows(区间DP)
2024-09-11 15:12:46
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:由里向外逆推区间
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std; int a[2005],dp[2005][2005]; int main()
{
int n,i,j,k,l,ans;
while(~scanf("%d",&n))
{
for(i = 1; i<=n; i++)
scanf("%d",&a[i]);
memset(dp,0,sizeof(dp));
for(i = 1; i<=n; i++)
dp[i][i] = a[i]*n;//将对角线初始化
for(l = 1; l<n; l++)
{
for(i = 1; i+l<=n; i++)
{
j = i+l;
dp[i][j] = max(dp[i+1][j]+(n-l)*a[i],dp[i][j-1]+(n-l)*a[j]);//这里是从最后出队的开始往前推,之前的初始化也是为了这里,因为只有最后出队的,i+1才会等于j。
}
}
printf("%d\n",dp[1][n]);
} return 0;
}
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