Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 51853    Accepted Submission(s): 21833 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 2 ³¹).   Sample Input 1 5 10 1 2 3 4 5 5 4 3 2 1   Sample Output 14 Author Teddy Source HDU 1st “Vegetable-Birds Cup” Programming Open Contest Recommend lcy   |   We have carefully selected several similar problems for you:   2159  2955  1171  2191  2844    01背包：

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int v[1100],w[1100],dp[1100]; int main() { int T; scanf("%d",&T); while(T--) { int N,V; scanf("%d%d",&N,&V); for(int i = 1; i <= N; i++) scanf("%d",&v[i]); for(int i = 1; i <= N; i++) scanf("%d",&w[i]); memset(dp,0,sizeof(dp)); for(int i = 1; i <= N; i++) { for(int j = V; j >= w[i]; j--) { dp[j] = max(dp[j],dp[j-w[i]] + v[i]); } } printf("%d\n",dp[V]); } return 0; }