Bone Collector

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 2602
64-bit integer IO format: %I64d Java class name: Main

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

Sample Output

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

解题：0-1背包。。。

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <cstdlib>

 #include <vector>

 #include <climits>

 #include <ctype.h>

 #include <cmath>

 #include <algorithm>

 #define LL long long

 using namespace std;

 int v[],w[],dp[];

 int main(){

     int kase,i,j,n,m;

     scanf("%d",&kase);

     while(kase--){

         memset(dp,,sizeof(dp));

         scanf("%d %d",&n,&m);

         for(i = ; i <= n; i++)

             scanf("%d",v+i);

         for(i = ; i <= n; i++)

             scanf("%d",w+i);

         for(i = ; i <= n; i++){

             for(j = m; j >= w[i]; j--){

                 dp[j] = max(dp[j],dp[j-w[i]]+v[i]);

             }

         }

         printf("%d\n",dp[m]);

     }

     return ;

 }

巴特西

BNUOJ 6727 Bone Collector