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  • hdoj

    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
     
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    
    using namespace std;
    int a[1005],b[1005],dp[1005];
    int main()
    {
        int t,n,v;
        scanf("%d",&t);
        for(int k=0;k<t;k++)
        {
            memset(dp,0,sizeof(dp));
            scanf("%d %d",&n,&v);
            for(int i = 0;i<n;i++)
            {
                scanf("%d",&a[i]);
            }
            for(int i = 0;i<n;i++)
            {
                scanf("%d",&b[i]);
            }
    
            for(int i = 0;i<n;i++)
            {
                for(int j = v;j>=b[i];j--)
                {
                    dp[j] = max(dp[j],dp[j-b[i]]+a[i]);      //关键代码
    
                }
            }
            printf("%d
    ",dp[v]);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/hdyss/p/10820486.html
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