HDU2602--Bone Collector--动态规划

 

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

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

#include <iostream>#include <cstdio>using namespace std;int dp[1008][1008];int value[1008];int vol[1008];int max(int a,int b){return a>b?a:b;}int main(){int t;scanf("%d",&t);while(t--){int n,v;memset(dp,0,sizeof(dp));scanf("%d%d",&n,&v);for(int i=1;i<=n;i++){scanf("%d",&value[i]);}for(int i=1;i<=n;i++){scanf("%d",&vol[i]);}for(int i=1;i<=n;i++)//石头个数{for(int j=0;j<=v;j++)//背包体积不断变大{dp[i][j]=dp[i-1][j];if(j>=vol[i]){dp[i][j]=max(dp[i][j],dp[i-1][j-vol[i]]+value[i]);}}}cout<<dp[n][v]<<endl;}return 0;}