HDU 2602 Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 41458    Accepted Submission(s): 17255

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

经典的01背包问题，本人博客其他文章有介绍，代码如下：

#include <stdio.h>#include <string.h>int dp[1005];int main(){int T,n;int i,j;int w[1005],v[1005],V;scanf("%d",&T);while(T--){memset(dp,0,sizeof(dp));scanf("%d %d",&n,&V);for(i=1;i<=n;i++)scanf("%d",v+i);for(j=1;j<=n;j++)scanf("%d",w+j);for(i=1;i<=n;i++){for(j=V;j>=w[i];j--){if(dp[j]<dp[j-w[i]]+v[i])dp[j]=dp[j-w[i]]+v[i];}}printf("%d\n",dp[V]);}return 0;}