Problem Q

Posted 2022-12-11 tansanity

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 …<br>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 ?<br><center><img src=../../../data/images/C154-1003-1.jpg> </center><br>  
Input The first line contain a integer T , the number of cases.<br>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

简单题意：   很久之前，有一个骨头收集者，当然不同的骨头拥有不同的价值和体积。现在他背着一个拥有V容量的包裹从小镇出发，在路途中给出骨头的价值。现在需要编写一个程序，是价值达到最大。输入有三行，第一行是骨头的数目和背包的容量V，第二行是骨头的价值，第三行是骨头的体积。输出最大的总价值。 解题思路形成过程：   这是典型的01背包问题，对于每一块骨头。可以选择放进包里或者不。01背包的状态转移方程为：dp[i][v]=maxdp[i-1][v],dp[i-1][v-cost[i]]+value[i]; 感想：   01背包上课刚讲完现在就有了用武之地。动态规划还是有很多子类别的。 AC代码： #include<iostream>
#include <cstring>
using namespace std;
#define Size 1001
int va[Size],vo[Size];
int dp[Size];
int Max(int x,int y)

    return x>y?x:y;

int main()

    int t,n,v,i,j;
    cin>>t;
    while(t--)
   
        cin>>n>>v;
        for(i=1;i<=n;i++)
            cin>>va[i];
        for(i=1;i<=n;i++)
            cin>>vo[i];
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++)
       
            for(j=v;j>=vo[i];j--)
           
                dp[j]=Max(dp[j],dp[j-vo[i]]+va[i]);
           
       
        cout<<dp[v]<<endl;
   
    return 0;