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Bone Collector Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 44243 Accepted Submission(s): 18434
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
Author Teddy
Source |
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 int dp[1010]; 6 struct node 7 { 8 int value; 9 int volume; 10 }m[1010]; 11 int main() 12 { 13 int t; 14 cin>>t; 15 int n,v; 16 while(t--){ 17 cin>>n>>v; 18 memset(dp,0,sizeof(dp)); 19 for(int i=1;i<=n;i++) cin>>m[i].value; 20 for(int i=1;i<=n;i++) cin>>m[i].volume; 21 for(int i=1;i<=n;i++){ 22 for(int j=v;j>=m[i].volume;j--){ 23 dp[j]=max(dp[j],dp[j-m[i].volume]+m[i].value); 24 } 25 } 26 cout<<dp[v]<<endl; 27 } 28 return 0; 29 }
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原文地址：http://www.cnblogs.com/shenyw/p/5184489.html
