01背包最簡單最經典的入門題

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

1
5 10
1 2 3 4 5
5 4 3 2 1 
             

Sample Output

14 
             

  題目大體意思是有n個東西，背包容量為m，給出每個東西的價值和體積，求背包能裝下的最大價值。有點簡單，不叨叨，直接給源代碼，都能看懂。

#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;
}