HDU2602 Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 24147    Accepted Submission(s): 9798

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

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

 

Author

Teddy

 

做的第一个背包题，个人感觉理解背包类问题解题方法（动态规划）是解题正确的关键

感觉此类问题与递推有较大的联系。一点拙见，不对之处请各位多多指教。代码如下:

 

 

#include<stdio.h>#include<string.h>int bone[1001];struct sa{    int W,V;}data[1001];int max(int a,int b){    return a>b?a:b;}int main(){    int t,i,j;    while(scanf("%d",&t)!=EOF)    {    while(t--)    {        memset(bone,0,sizeof(bone));        int n,v;        scanf("%d%d",&n,&v);        for(i=1;i<=n;i++)        scanf("%d",&data[i].W);        for(j=1;j<=n;j++)        scanf("%d",&data[j].V);        for(i=1;i<=n;i++)        for(j=v;j>=data[i].V;j--)        {            bone[j]=max(bone[j],bone[j-data[i].V]+data[i].W);        }        printf("%d\n",bone[v]);    }    }    return 0;}