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O - Treats for the Cows 区间DP
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 first treat is sold on day 1 and has age a=1. Each subsequent day increases the age by 1.
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.
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)
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.
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.
注意从小区间推大区间,从内向外推,也就是从后卖出的物品向前卖出的物品状态递推。
#include <stdio.h> #include <string.h> #include <algorithm> using namespace std; #define MAXN 2002 /* 区间DP,从后向前推,dp[i][j]表示首端元素为a[i],尾端为a[j]的情况 dp[i][j] = max(dp[i+1][j]+t*a[i],dp[i][j-1]+t*a[j]) */ int a[MAXN],dp[MAXN][MAXN]; int main() { int n; scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&a[i]); for(int i=1;i<=n;i++) dp[i][i] = n*a[i];//最后一个卖出元素是a[i]的情况 for(int l=1;l<n;l++) { for(int i=1;i+l<=n;i++) { int j = i+l; dp[i][j] = max(dp[i+1][j]+(n-l)*a[i],dp[i][j-1]+(n-l)*a[j]); } } printf("%d\n",dp[1][n]); return 0; }
O - Treats for the Cows 区间DP
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