hdu 2602 Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 52158    Accepted Submission(s): 21977

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

 

Author

Teddy

标准的0-1背包 下面的代码我试了  可以ac！

这是小编写的第一个背包题，说说心得，就是用二维数组保存所有情况的最优，然后，，，就出来了，第一次写，理解的不对的 ，求大神指出，谢谢

#include <stdio.h>#include <string.h>int dp[1005][1005];int max(int x,int y){return x>y?x:y;}int main(){int n,v,t,i,j;int va[1005],vo[1005];while(~scanf("%d",&t)){while(t--){scanf("%d%d",&n,&v);for(i=1;i<=n;i++){scanf("%d",&va[i]);}for(i=1;i<=n;i++){scanf("%d",&vo[i]);}memset(dp,0,sizeof(dp));for(i=1;i<=n;i++){for(j=0;j<=v;j++){if(vo[i]<=j)dp[i][j]=max(dp[i-1][j],dp[i-1][j-vo[i]]+va[i]);elsedp[i][j]=dp[i-1][j];}}printf("%d\n",dp[n][v]);}}return 0;}