HDU 2602 Bone Collector

Bone Collector

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 2602

Appoint description:  System Crawler  (2017-04-13)

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 2 ³¹).

 

Sample Input

1

5 10

1 2 3 4 5

5 4 3 2 1

 

Sample Output

14

 一个简单的01背包问题

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
long long a[1010],v[1010];
long long dp[1010];
int n,res;
long long max(long long a,long long b)
{
    if(a>=b)
        return a;
    else return b;
}
int main()
{
    int T;
    while( cin >> T )
    {
        while( T -- )
        {
            cin >> n >> res;
            for(int i=0; i<n; i++)
                cin >> v[i];
            for(int j=0; j<n; j++)
                cin >> a[j];
            memset(dp,0,sizeof(dp));//初始化
            for(int i=0; i<n; i++)
            {
                for(int j=res; j>=a[i]; j--)
                {
                    dp[j] = max(dp[j], dp[j-a[i]]+v[i]);//01背包，理解
                }
            }
            cout << dp[res] << endl;
        }
    }
    return 0;
}