Problem Q

最新推荐文章于 2019-10-28 00:12:41 发布

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本文介绍了一个经典的01背包问题实例，通过一个骨骸收集者的故事情境来阐述如何使用动态规划解决该问题。文章提供了详细的解题思路及AC代码实现。

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 ? <center><img src=../../../data/images/C154-1003-1.jpg> </center>

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

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]=max{dp[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;
}