HDUOJ--Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21463    Accepted Submission(s): 8633

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 2 ³¹).

 

 

Sample Input

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

 

 

Sample Output

14

 

 

Author

Teddy

 

 

Source

 

  http://acm.hdu.edu.cn/showproblem.php?pid=2602

 

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代码：----

 1 #include<stdio.h>

 2 #include<string.h>

 3 #define maxn 1005

 4 int dp[maxn],arr[maxn][2];

 5 int max(int a,int b)

 6 {

 7     return a>b?a:b;

 8 }

 9 

10 void zeroonepack(int cost ,int value,int v)

11 {

12      for(int i=v;i>=cost;i--)

13          dp[i]=max(dp[i],dp[i-cost]+value);

14 

15 }

16 int main()

17 {

18     int t,n,v ,i;

19     scanf("%d",&t);

20     while(t--)

21     {

22         scanf("%d%d",&n,&v);

23         memset(dp,0,sizeof dp);

24         for(i=0;i<n;i++)

25         {

26             scanf("%d",arr[i]+0);

27         }

28         for(i=0;i<n;i++)

29         {

30             scanf("%d",arr[i]+1);

31         }

32        for(i=0;i<n;i++)

33            zeroonepack(arr[i][1],arr[i][0],v);

34        printf("%d\n",dp[v]);

35     }

36     return 0;

37 }

View Code

 优化后代码：

#include<stdio.h>

#include<stdlib.h>

#include<string.h>

struct st

{

    int a;

    int b;

};

typedef struct st sta;

int main()

{

    int test,n,v,i,j;

    scanf("%d",&test);

    while(test--)

    {

        scanf("%d%d",&n,&v);

     int *dp =(int *)malloc(sizeof(int)*(v+1));

     sta *stu =(sta *)malloc(sizeof(sta)*(n+1));

        for(i=0;i<n;i++)

            scanf("%d",&stu[i].a);

        for(i=0;i<n;i++)

            scanf("%d",&stu[i].b);

        for(i=0;i<=v;i++)

            dp[i]=0;



       for(i=0;i<n;i++)

       {

           for(j=v ; j>=stu[i].b ; j--)

           {

            if(dp[j]<dp[j-stu[i].b]+stu[i].a)

               dp[j]=dp[j-stu[i].b]+stu[i].a;

           }

       }

       printf("%d\n",dp[v]);

       free(dp);

       free(stu);

    }

    return 0;

}