HDU 2602 Bone Collector【01DP（二）】

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 52273    Accepted Submission(s): 22025

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

 

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<climits>#include<string>#include<queue>#include<stack>#include<set>#include<map>#include<algorithm>using namespace std;#define rep(i,j,k)for(i=j;i<k;i++)#define per(i,j,k)for(i=j;i>k;i--)#define MS(x,y)memset(x,y,sizeof(x))typedef long long LL;const int INF=0x7ffffff;const int M=1e3+1;int val[M],vol[M];int dp[M];int i,j,k,n,m,t;int main(){   int T;   scanf("%d",&T);   while(T--)   {       scanf("%d%d",&n,&m);       MS(dp,0);       rep(i,0,n)scanf("%d",&val[i]);       rep(i,0,n)scanf("%d",&vol[i]);       rep(i,0,n)         for(j=m;j>=vol[i];j--)            dp[j]=max(dp[j],dp[j-vol[i]]+val[i]);         printf("%d\n",dp[m]);   }   return 0;}