HDU 2602 Bone Collector
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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
15 101 2 3 4 55 4 3 2 1
Sample Output
14经典01背包问题f[j]=max(f[j],f[j-cost[i]]+get[i])LANGUAGE:CCODE:#include<stdio.h>#define max(a,b) a>b?a:bint cost[1002],get[1002],f[1002];int main(){ int testCase,N,V,i,j; scanf("%d",&testCase); while(testCase--) { scanf("%d%d",&N,&V); for(i=0;i<=V;i++)f[i]=0; for(i=1;i<=N;i++) scanf("%d",&get[i]); for(i=1;i<=N;i++) scanf("%d",&cost[i]); for(i=1;i<=N;i++) for(j=V;j>=cost[i];j--) f[j]=max(f[j],f[j-cost[i]]+get[i]); printf("%d\n",f[V]); } return 0;}
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