HDU2602:Bone Collector(01背包)

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 <stdio.h>#include <string.h>struct Node{    int h;    int v;}node[1005];int max(int a,int b){    return a>b?a:b;}int main(){    int t,n,m,l;    int dp[1005];    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        int i;        for(i = 1;i<=n;i++)        scanf("%d",&node[i].h);        for(i = 1;i<=n;i++)        scanf("%d",&node[i].v);        memset(dp,0,sizeof(dp));        for(i = 1;i<=n;i++)        {            for(l = m;l>=node[i].v;l--)            dp[l] = max(dp[l],dp[l-node[i].v]+node[i].h);        }        printf("%d\n",dp[m]);    }    return 0;}