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 2³¹).

 

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。。。

 

#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[1001],w[1001],dp[1001];
int main(){
    int kase,i,j,n,m;
    scanf("%d",&kase);
    while(kase--){
        memset(dp,0,sizeof(dp));
        scanf("%d %d",&n,&m);
        for(i = 1; i <= n; i++)
            scanf("%d",v+i);
        for(i = 1; i <= n; i++)
            scanf("%d",w+i);
        for(i = 1; i <= n; i++){
            for(j = m; j >= w[i]; j--){
                dp[j] = max(dp[j],dp[j-w[i]]+v[i]);
            }
        }
        printf("%d
",dp[m]);
    }
    return 0;
}