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Bone Collector(杭电2602)(01背包)

2014-11-08 22:23:00.0 动态规划  
导读:Bone CollectorTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 31604    Accepted Submission(s): 13005Problem DescriptionMany years ago , in Teddy...。。。

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 31604    Accepted Submission(s): 13005


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
/*简单背包问题*/#include#include#includeusing namespace std;int main(){ int test,n,v,i,j; int a[1100],s[1100],dp[1100]; scanf("%d",&test); while(test--){  memset(dp,0,sizeof(dp)); scanf("%d %d",&n,&v); for(i=1;i<=n;i++) {  scanf("%d",&a[i]); } for(i=1;i<=n;i++) {  scanf("%d",&s[i]); } for(i=1;i<=n;i++) {  for(j=v;j>=s[i];j--)  {   dp[j]=max(dp[j],dp[j-s[i]]+a[i]);  } } printf("%d\n",dp[v]);} return 0;} 


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