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  • hdu 2602

    Time Limit: 1000MS   Memory Limit: 32768KB   64bit IO Format: %I64d & %I64u

     Status

    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 2 31).
     

    Sample Input

    1 5 10 1 2 3 4 5 5 4 3 2 1
     

    Sample Output

    14
     

    Source

    HDU 1st “Vegetable-Birds Cup” Programming Open Contest
     
     
    01背包
    #include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <cstdio>
    #include <cmath>
    
    using namespace std;
    
        int T;
        int n,v;
        const int N=1E3+5;
        int a[N],b[N],dp[N];
    
    void ZeroOnePack(int value, int cost)
    {
        for ( int i = v ; i >= cost ; i-- )
            dp[i] = max(dp[i],dp[i-cost]+value);
    }
    
    int main()
    {
    
        scanf("%d",&T);
        while ( T-- )
        {  memset(dp,0,sizeof(dp));
    
            scanf("%d %d",&n,&v);
            for ( int i = 1; i <= n ; i++ )
                scanf("%d",&a[i]);
            for ( int i = 1 ; i <= n ; i++ )
                scanf("%d",&b[i]);
            for ( int i = 1 ; i <= n ; i++ )
                ZeroOnePack(a[i],b[i]);
            printf("%d
    ",dp[v]);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/111qqz/p/4392650.html
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