杭电2602

哎，动态规划背包搞了好多天，终于搞懂了。。。坚持就是胜利

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 43080    Accepted Submission(s): 17962

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

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

 

#include<iostream>using namespace std;#define MAXN 1000int d[MAXN][MAXN];int main(){int t,n,m,a[MAXN],b[MAXN],i,j;while(cin>>t){while(t--){cin>>n>>m;for(i=1;i<=n;i++)cin>>a[i];for(j=1;j<=n;j++)cin>>b[j];for(i=1;i<=n;i++){for(j=0;j<=m;j++){if(j>=b[i])         //第i个物品装入背包 {d[i][j]=d[i-1][j]>(d[i-1][j-b[i]]+a[i])?d[i-1][j]:(d[i-1][j-b[i]]+a[i]);  }else        //第i个物品不装入 d[i][j]=d[i-1][j];}}cout<<d[n][m]<<endl;}}return 0;}