HDU 2602 Bone Collector

E - Bone Collector

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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 2 ³¹).

 

Sample Input

 15 101 2 3 4 55 4 3 2 1 

 

Sample Output

 14 
一个简单的01背包
#include<stdio.h>#include<string.h>#define MAX 100001int dp[MAX],w[MAX],p[MAX];int max(int a,int b){    return a > b? a:b;}int main(){    int T,i,j;    int n,m;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&m);        for(i=1;i<=n;i++)        {            scanf("%d",&w[i]);        }        for(i=1;i<=n;i++)        {            scanf("%d",&p[i]);        }        memset(dp,0,sizeof(dp));        for(i=1;i<=n;i++)        {            for(j=m;j>=p[i];j--)            {                dp[j] = max(dp[j], dp[j-p[i]]+w[i]);            }        }        printf("%d\n",dp[m]);    }    return 0;}