HDU 2602Bone Collector（01背包）

Bone Collector

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

Total Submission(s) : 4   Accepted Submission(s) : 1

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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

总结：裸裸的01背包

////  main.cpp//  Bone Collector////  Created by 张嘉韬 on 16/8/8.//  Copyright © 2016年 张嘉韬. All rights reserved.//#include <iostream>#include <cstring>#include <cstdio>#include <cmath>using namespace std;const int maxn=1000+10;int f[maxn][maxn],w[maxn],v[maxn];int main(int argc, const char * argv[]) {   // freopen("/Users/zhangjiatao/Documents/暑期训练/input.txt","r",stdin);    int T;    cin>>T;    while(T--)    {        int n,p;        memset(f,0,sizeof(f));        scanf("%d%d",&n,&p);        for(int i=1;i<=n;i++) scanf("%d",&v[i]);        for(int i=1;i<=n;i++) scanf("%d",&w[i]);        for(int i=1;i<=n;i++)        {            for(int j=0;j<=p;j++)            {                if(j-w[i]<0) f[i][j]=f[i-1][j];                else f[i][j]=max(f[i-1][j],f[i-1][j-w[i]]+v[i]);            }        }        int maximum=0;        for(int i=0;i<=p;i++) if(f[n][i]>maximum) maximum=f[n][i];        cout<<maximum<<endl;    }    return 0;}