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  • hdu 2602 Bone Collector (01背包)

    Bone Collector

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 27002    Accepted Submission(s): 10938


    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
    Teddy
     
    Source
     
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    01背包模板题:

     1 //15MS    240K    545 B    C++
     2 #include<stdio.h>
     3 #include<string.h>
     4 int dp[1005];
     5 int v[1005],w[1005];
     6 int Max(int a,int b)
     7 {
     8     return a>b?a:b;
     9 }
    10 int main(void)
    11 {
    12     int t,n,m;
    13     scanf("%d",&t);
    14     while(t--)
    15     {
    16         scanf("%d%d",&n,&m);
    17         memset(dp,0,sizeof(dp));
    18         for(int i=1;i<=n;i++) scanf("%d",&v[i]);
    19         for(int i=1;i<=n;i++) scanf("%d",&w[i]);
    20         for(int i=1;i<=n;i++)
    21             for(int j=m;j>=w[i];j--)
    22                 dp[j]=Max(dp[j],dp[j-w[i]]+v[i]);
    23         printf("%d
    ",dp[m]);
    24     }
    25     return 0;
    26 } 
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  • 原文地址:https://www.cnblogs.com/GO-NO-1/p/3740502.html
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