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HDU2602Bone Collector

2024-07-16 21:19:38   221人阅读

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29685    Accepted Submission(s): 12208


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
 
简单01背包
#include<iostream>#include<cstdio>#include<cmath>#include<queue>#include<algorithm>#include<cstring>using namespace std;int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,i,j,v;        scanf("%d%d",&n,&v);        int dp[1100]={0},c[1100],w[1100];        for(i=0;i<n;i++)            scanf("%d",w+i);        for(i=0;i<n;i++)            scanf("%d",c+i);        for(i=0;i<n;i++)        {            for(j=v;j>=c[i];j--)            {                if(dp[j]<dp[j-c[i]]+w[i])                {                    dp[j]=dp[j-c[i]]+w[i];                }            }        }        printf("%d\n",dp[v]);    }    return 0;}


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