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  • hdu 2602(01背包)

    Bone Collector

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 46546    Accepted Submission(s): 19378


    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
     
    Author
     
    水题.
    #include<stdio.h>
    #include<algorithm>
    #include<string.h>
    #include<iostream>
    #define N 1005
    using namespace std;
    
    int dp[N];
    int V[N],W[N];
    int main()
    {
        int tcase;
        scanf("%d",&tcase);
        while(tcase--){
            int n,m;
            scanf("%d%d",&n,&m);
            for(int i=1;i<=n;i++){
                scanf("%d",&W[i]);
            }
            for(int i=1;i<=n;i++){
                scanf("%d",&V[i]);
            }
            memset(dp,0,sizeof(dp));
            for(int i=1;i<=n;i++){
                for(int v=m;v>=V[i];v--){
                    dp[v] = max(dp[v],dp[v-V[i]]+W[i]);
                }
            }
            printf("%d
    ",dp[m]);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/liyinggang/p/5392715.html
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