hdu 2602(01背包)

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 46546    Accepted Submission(s): 19378

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

1 5 10 1 2 3 4 5 5 4 3 2 1

 

Sample Output

14

 

Author

 

水题.

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
#define N 1005
using namespace std;

int dp[N];
int V[N],W[N];
int main()
{
    int tcase;
    scanf("%d",&tcase);
    while(tcase--){
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++){
            scanf("%d",&W[i]);
        }
        for(int i=1;i<=n;i++){
            scanf("%d",&V[i]);
        }
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++){
            for(int v=m;v>=V[i];v--){
                dp[v] = max(dp[v],dp[v-V[i]]+W[i]);
            }
        }
        printf("%d
",dp[m]);
    }
    return 0;
}