BNUOJ 6727 Bone Collector

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 2 ³¹).

 

Sample Input

1

5 10

1 2 3 4 5

5 4 3 2 1

 

Sample Output

14

 

Source

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

 

 

解题：0-1背包。。。

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cstdlib>

 5 #include <vector>

 6 #include <climits>

 7 #include <ctype.h>

 8 #include <cmath>

 9 #include <algorithm>

10 #define LL long long

11 using namespace std;

12 int v[1001],w[1001],dp[1001];

13 int main(){

14     int kase,i,j,n,m;

15     scanf("%d",&kase);

16     while(kase--){

17         memset(dp,0,sizeof(dp));

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

19         for(i = 1; i <= n; i++)

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

21         for(i = 1; i <= n; i++)

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

23         for(i = 1; i <= n; i++){

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

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

26             }

27         }

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

29     }

30     return 0;

31 }

View Code