HDU2602 01背包问题

最新推荐文章于 2020-08-26 16:53:54 发布

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本文介绍了一个经典的背包问题案例，通过编程解决如何使骨收集者在有限的背包容量下获得最大价值的问题。输入包括测试案例数量、骨头数量及背包容量、每根骨头的价值和体积等信息。输出则是最大总价值。

挺想哭的调试了好久都没用，结果是int的问题。。。

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

#include<iostream>
#include<stdio.h>
#include<memory.h>
using namespace std;
int mon[1003],w[1003];
int f[1001]={0};
int max(int x,int y)
{
        if(x>y)
                return x;
        else
                return y;
}
int main()
{
        int n,v,i,j,k;
        int t;
        cin>>t;
        for( k=0;k<t;k++)
        {
                memset(f,0,sizeof(f));
                cin>>n>>v;
                for(i=0;i<n;i++)
                {
                        cin>>mon[i];
                }
                 for(i=0;i<n;i++)
                {
                        cin>>w[i];
                }
                //for(i=0;i<n;i++)
                // {
                //         cout<<mon[i]<<w[i];
                // }
                for(i=0;i<n;i++)
                {
                        for(j=v;j>=w[i];j--)
                        {
                              f[j]=max(f[j],f[j-w[i]]+mon[i]);
                              //cout<<f[v]<<endl;
                        }
                }
                cout<<f[v]<<endl;
        }
        /*for(i=0;i<t;i++)
        {
                cout<<a[i]<<endl;
        }*/
        return 0;
}