2016SDAU课程练习三1017 Problem Q

Problem Q

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 130   Accepted Submission(s) : 52

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 …<br>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 ?<br><center><img src=../../../data/images/C154-1003-1.jpg> </center><br>

 

Input

The first line contain a integer T , the number of cases.<br>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

题目大意：

给出袋子的最大容量，和每个物品的价值和体积，求最多能装多少价值的物品。

思路：

就是简单的​01背包问题，照着例题很快可以编出。

ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int volume[1005];
int value[1005];
int dp[1005];
int n,m;

int main(){

    int T;
    scanf("%d",&T);
    while(T--){
        int N,V;
        memset(dp,0,sizeof(dp));
        scanf("%d %d",&N,&V);
        for(int i=1;i<=N;i++)
            scanf("%d",&value[i]);
        for(int i=1;i<=N;i++)
            scanf("%d",&volume[i]);
        for(int i=1;i<=N;i++)
            for(int j=V;j>=volume[i];j--)
                dp[j]=max(dp[j],dp[j-volume[i]]+value[i]);
        printf("%d\n",dp[V]);
    }
    return 0;
}