Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 48618 Accepted Submission(s): 20269
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
注意：先输入每个骨头的价值然后输入每个骨头的体积。
状态转移方程：dp[j]=max(dp[j],dp[j-v[i]]+w[i]。
#include <iostream> #include <string.h>using namespace std; int N,V; int dp[1111]; int w[1111],v[1111]; int main() {int t;while(cin>>t){while(t--){cin>>N>>V;for(int i=1;i<=N;i++){cin>>w[i];}for(int i=1;i<=N;i++){cin>>v[i];}memset(dp,0,sizeof(dp));for(int i=1;i<=N;i++){for(int j=V;j>=v[i];j--){ //j如果小于了v[i]那么v[i]一定无法装入袋子dp[j]=max(dp[j],dp[j-v[i]]+w[i]);}}cout<<dp[V]<<endl;}}return 0; }
