Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 21417 Accepted Submission(s): 8613
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
1
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14
#include<iostream> #include<cstring> #include<string> using namespace std; int main() { int t; scanf("%d",&t); while(t--) { int n,m,i,j,value[1020],volume[1020],dp[1020]; scanf("%d %d",&n,&m); for(i=0;i<n;i++) { scanf("%d",&value[i]); } for(i=0;i<n;i++) { scanf("%d",&volume[i]); } memset(dp,0,sizeof(dp)); for(i=0;i<n;i++) { for(j=m;j>=volume[i];j--) { dp[j] = dp[j-volume[i]] + value[i] > dp[j] ? dp[j-volume[i]]+value[i] : dp[j]; //if((dp[j-v])) } } printf("%d ",dp[m]); } return 0; }