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Hdu oj 2602 Bone Collector(01背包)

热度:13   发布时间:2023-12-22 04:14:43.0

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 56560    Accepted Submission(s): 23661


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 2 31).

Sample Input
  
   
1 5 10 1 2 3 4 5 5 4 3 2 1

Sample Output
  
   
14
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;const int Maxn = 1005;
int v[Maxn];
int w[Maxn];
int dp[Maxn];
int N,V;
int t;int main()
{scanf("%d",&t);while(t--){scanf("%d%d",&N,&V);for(int i=1;i<=N;i++){scanf("%d",&v[i]);}for(int i=1;i<=N;i++){scanf("%d",&w[i]);}memset(dp,0,sizeof(dp));for(int i=1;i<=N;i++){for(int j=V;j>=w[i];j--){dp[j] = max(dp[j],dp[j-w[i]] + v[i]);}}printf("%d\n",dp[V]);}
}