Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 40316 Accepted Submission(s): 16748
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<stdio.h> #include<string.h> #include<algorithm> using namespace std; struct stu { int val; int cos; }boy[10010]; //stu ; //int bag[10010]; int nexty[10010];//如果把所有nexty都改为next提交时出现编译错误。 int main() { int n; int a,b; scanf("%d",&n); while(n--) { scanf("%d%d",&a,&b); memset(boy,0,sizeof(boy)); // memset(bag,0,sizeof(bag)); memset(nexty,0,sizeof(nexty)); for(int i=0;i<a;i++) scanf("%d",&boy[i].val); for(int j=0;j<a;j++) scanf("%d",&boy[j].cos); for(int i=0;i<a;i++) { for(int j=b;j>=boy[i].cos;j--) // bag[j]=max(bag[j],bag[j-boy[i].cos]+boy[i].val); nexty[j]=max(nexty[j],nexty[j-boy[i].cos]+boy[i].val); } //printf("%d\n",bag[b]); printf("%d\n",nexty[b]); } return 0; }
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