【01背包】HDU 2602 Bone Collector （模板题）

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

Sample Output

Author

Teddy

Source

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

思路：01背包裸题。01背包：有限的背包容量取最大的价值，每样东西选择取或不取

 #include <iostream>

 #include<cstdio>

 #include<algorithm>

 #include<cstring>

 #include<cmath>

 using namespace std;

 long long va[];

 int vo[];

 long long dp[];

 int main()

 {

     int T,n,v,i,j;

     scanf("%d",&T);

     while(T--)

     {

         scanf("%d%d",&n,&v);

         for(i=;i<=n;i++)

             scanf("%lld",&va[i]);

         for(i=;i<=n;i++)

             scanf("%d",&vo[i]);

         memset(dp,,sizeof(dp));

         for(i=;i<=n;i++)

         {

             for(j=v;j>=vo[i];j--)

             {

                 dp[j]=max(dp[j],dp[j-vo[i]]+va[i]);

             }

         }

         printf("%lld\n",dp[v]);

     }

     return ;

 }

巴特西