HDU 2602 Bone Collectors(背包问题，模版）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14336    Accepted Submission(s): 5688

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

Teddy

 

Source

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

 

Recommend

lcy

这道题，我学会了，dp数组一定要放在主函数的外面，不然编译器就会崩！！！

这里的j起始要从0开始，这是十分恶心的地方，肯能就是volume=0也会有value！！！这个是个巨型的坑！！！！！

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[][];
int main()
{
    int t;
    cin>>t;
    int n,m;
    int w[],v[];
    int f1,f2;
    while(t--)
    {
        cin>>n>>m;
        memset(w,,sizeof(w));
        memset(v,,sizeof(v));
        memset(dp,,sizeof(dp));
        for(int i=;i<=n;i++)
        {
            cin>>v[i];
        }
        for(int i=;i<=n;i++)
        {
            cin>>w[i];
        }
        for(int i=;i<=n;i++)
        {
            for(int j=;j<=m;j++)
            {
                if(j<w[i]) dp[i][j]=dp[i-][j];
                else
                {
                    f1=i;
                    f2=j;
                    dp[i][j]=max(dp[i-][j],dp[i-][j-w[i]]+v[i]);
                }
            }
        }
        cout<<dp[f1][f2]<<endl;
    }
    return ;
}