杭电-2602 Bone Collector（01背包模板）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 51853    Accepted Submission(s): 21833

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

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

Author

Teddy

Source

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

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01背包：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int v[1100],w[1100],dp[1100];int main(){    int T;    scanf("%d",&T);    while(T--)    {        int N,V;        scanf("%d%d",&N,&V);        for(int i = 1; i <= N; i++)            scanf("%d",&v[i]);        for(int i = 1; i <= N; i++)            scanf("%d",&w[i]);        memset(dp,0,sizeof(dp));        for(int i = 1; i <= N; i++)        {            for(int j = V; j >= w[i]; j--)            {                dp[j] = max(dp[j],dp[j-w[i]] + v[i]);            }        }        printf("%d\n",dp[V]);    }    return 0;}