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 ?
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.
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> using namespace std; int volume[1005]; int value[1005]; int dp[1005]; int n,m; int main() { freopen("C:\\in.txt","r",stdin); int T; scanf("%d",&T); while(T--){ int N,V; memset(dp,0,sizeof(dp)); scanf("%d %d",&N,&V); for(int i=1;i<=N;i++) scanf("%d",&value[i]); for(int i=1;i<=N;i++) scanf("%d",&volume[i]); for(int i=1;i<=N;i++) for(int j=V;j>=volume[i];j--) dp[j]=max(dp[j],dp[j-volume[i]]+value[i]); printf("%d\n",dp[V]); } return 0; }