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  • HDU 2602 Bone Collector (简单01背包)

    Bone Collector

    http://acm.hdu.edu.cn/showproblem.php?pid=2602

    Problem Description
    Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
    The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?
     
    Input
    The first line contain a integer T , the number of cases.
    Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.
     
    Output
    One integer per line representing the maximum of the total value (this number will be less than 231).
     
    Sample Input
    1
    5 10
    1 2 3 4 5
    5 4 3 2 1
     
    Sample Output
    14
     

    解题代码:

    #include <stdio.h>
    #include <math.h>
    #include <string.h>
    #include <iostream>
    using namespace std;
    
    const int max_v = 1004;
    int dp[max_v];
    struct boot
    {
        int val, cost;    
    };
    
    int main()
    {
        int T;
        scanf ("%d", &T);
        int N, V;
        boot B[max_v];
        while (T--)
        {
            memset (dp, 0, sizeof (dp));
            scanf ("%d%d", &N, &V);
            for (int i = 1; i <= N; i ++)
                scanf ("%d", &B[i].val);
            for (int i = 1; i <= N; i ++)
                scanf ("%d", &B[i].cost);
            for (int i = 1; i <= N; i ++)
            {
                for (int v = V; v >= B[i].cost; v --)
                    dp[v] = max(dp[v], dp[v-B[i].cost] + B[i].val);            
            }
            printf ("%d
    ", dp[V]);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/shengshouzhaixing/p/3195891.html
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