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  • (区间dp + 记忆化搜索)Treats for the Cows (POJ 3186)

     

    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<1 || R<1 || R>n || L>n) return -1;
    
        if(dp[L][R]!=-1)
            return dp[L][R];
    
        dp[L][R] = 0;
        dp[L][R] = max(DFS(L+1, R, k+1) + a[L]*k, DFS(L, R-1, k+1) + a[R]*k);
    
        return dp[L][R];
    }
    
    int main()
    {
    
        while(scanf("%d", &n)!=EOF)
        {
            int i;
    
            met(dp, -1);
            met(a, 0);
    
            for(i=1; i<=n; i++)
                scanf("%d", &a[i]);
    
            for(i=1; i<=n; i++)
                dp[i][i] = a[i]*n;
    
            dp[1][n] = DFS(1, n, 1);
    
            printf("%d
    ", dp[1][n]);
        }
        return 0;
    }

     

     

    #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, 0);
            met(a, 0);
    
            for(i=1; i<=n; i++)
                scanf("%d", &a[i]);
    
            for(i=1; i<=n; i++)
                dp[i][i] = a[i]*n;
    
            for(l=1; l<n; l++)
            {
                for(i=1; i+l<=n; i++)
                {
                    j = i+l;
                    dp[i][j] = max(dp[i+1][j]+a[i]*(n-l), dp[i][j-1]+a[j]*(n-l));
                }
            }
    
            printf("%d
    ", dp[1][n]);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/YY56/p/5469030.html
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