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
思路如下:
这题是一个标准的01背包问题,我们需要理解01背包的状态转移方程。
题解如下
#include<iostream>#include<cstring>#include<algorithm>
using namespace std;constint Len =1005;int dp[Len];int v[Len],w[Len];//v物品所占的空间,w品的价值intmain(){int t;
cin>>t;while(t --){int n,m;
cin>>n>>m;for(int i =0;i < n;i ++)
cin>>w[i];for(int i =0;i < n;i ++)
cin>>v[i];memset(dp,0,sizeof(dp));for(int i =0;i < n;i ++)for(int j = m;j >= v[i];j --)
dp[j]=max(dp[j],dp[j - v[i]]+ w[i]);
cout<<dp[m]<<endl;}return0;}