Bone Collector
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<cstdio> #include<cstring> #include<algorithm> using namespace std; int pa[1011];//价值 int st[1011];//重量 int dp[1011];//当前背包各种重量下的最优解 int main() { int T; int N,V; int i,j; scanf("%d",&T); while(T--) { scanf("%d%d",&N,&V); memset(dp,0,sizeof(dp)); for(i=1;i<=N;i++) scanf("%d",&pa[i]); for(i=1;i<=N;i++) scanf("%d",&st[i]); for(i=1;i<=N;i++) { for(j=V;j>=st[i];j--) { dp[j]=max(dp[j],dp[j-st[i]]+pa[i]);//当前背包各种重量下的最优解 } } printf("%d\n",dp[V]);//输出容量为V时的最优解 } return 0; }