Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 51789    Accepted Submission(s): 21810 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  2844  1114   

01背包问题

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; struct node { int num,val; }t[110000]; int main() { int i,j,k,l,T,v,n,m,dp[1100]; scanf("%d",&T); while(T--) { memset(dp,0,sizeof(dp)); scanf("%d%d",&n,&v); for(i=0;i<n;i++) scanf("%d",&t[i].val); for(i=0;i<n;i++) scanf("%d",&t[i].num); for(i=0;i<n;i++) for(j=v;j>=t[i].num;j--) { dp[j]=max(dp[j],dp[j-t[i].num]+t[i].val); } printf("%d\n",dp[v]); } return 0; }