背包问题 入门 hdu 2602

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21557    Accepted Submission(s): 8676

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

 

 

 

 01背包问题的入门题目

 

 

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int dp[1003];int p[1003];int v[1003];inline int max(int a,int b){    return a>b?a:b;}int  main(){    int i,j,k,t,n,V;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&V);        for(i=0;i<n;i++)        scanf("%d",&p[i]);        for(i=0;i<n;i++)        scanf("%d",&v[i]);        memset(dp,0,sizeof(dp));        for(i=0;i<n;i++)        {            for(j=V;j>=v[i];j--)            {                 dp[j]=max(dp[j],dp[j-v[i]]+p[i]);            }        }        printf("%d\n",dp[V]);    }    return 0;}