POJ3186——Treats for the Cows

2891 단어 dp
Treats for the Cows
Time Limit: 1000MS
 
Memory Limit: 65536K
Total Submissions: 4234
 
Accepted: 2132
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.
    Source
    USACO 2006 February Gold & Silver
    구간 dp, [i, j]를 고려하고 다른 것을 아직 꺼내지 않았다고 가정하면 [i, j]는 반드시 [i+1, j] 또는 [i, j-1]에서 추측할 수 있다.
    dp[i][j] = max(dp[i + 1][j] + v[i] * (n - (j - i)), dp[i][j - 1] + v[j] * (n - (j - i)))
    #include <map>
    #include <set>
    #include <list>
    #include <queue>
    #include <stack>
    #include <vector>
    #include <cmath>
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    
    using namespace std;
    
    const int N = 2010;
    int dp[N][N];
    int val[N];
    
    int main()
    {
    	int n;
    	while(~scanf("%d", &n))
    	{
    		memset (dp, 0, sizeof(dp));
    		for (int i = 1; i <= n; ++i)
    		{
    			scanf("%d", &val[i]);
    			dp[i][i] = val[i];
    		}
    		for (int i = n; i >= 1; --i)
    		{
    			for (int j = i; j <= n; ++j)
    			{
    				dp[i][j] = max(dp[i + 1][j] + val[i] * (n - (j - i + 1) + 1), dp[i][j - 1] + val[j] *(n - (j - i + 1) + 1) );
    			}
    		}
    		printf("%d
    ", dp[1][n]); } return 0; }

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