hdu 2602 Bone Collector

Bone Collector

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 56   Accepted Submission(s) : 36

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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

15 101 2 3 4 55 4 3 2 1

Sample Output

14

一道十分简单的01背包果题  

#include <cstdio>  #include <cstring>#include<iostream>  #include <algorithm>  using namespace std;  struct st{    int zl;    int jz;};int dp[1111];int main()  {  int k;cin>>k;while(k--){    int n,m;    memset(dp,0,sizeof(dp));    cin>>n>>m;    int i,j;    st st1[1111];    for(i=1;i<=n;i++)    scanf("%d",&st1[i].jz);    for(i=1;i<=n;i++)    scanf("%d",&st1[i].zl);    for(i=1;i<=n;i++)    for(j=m;j>=st1[i].zl;j--)    {        dp[j]=max(dp[j],dp[j-st1[i].zl]+st1[i].jz);    }    printf("%d\n",dp[m]);        }     return 0;  }