HDOJ 2602

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14134    Accepted Submission(s): 5585

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

View Code

 1 //下面的不对 

 2 #include <iostream>

 3 #include <cstring>

 4 using namespace std;

 5 typedef struct Node

 6 {

 7     int w,value;

 8 }Node;

 9 Node bag[1010];

10 int f[1010][1010];

11 int main()

12 {

13     int i,j,k,t,T;

14     cin>>T;

15     while(T--)

16     {

17         int n,c;

18         cin>>n>>c;

19         memset(bag,0,sizeof(bag));

20         memset(f,0,sizeof(f));

21         for(i=1;i<=n;i++)

22             cin>>bag[i].value;

23         for(i=1;i<=n;i++)

24             cin>>bag[i].w;

25         f[1][c] = bag[1].value;

26         for(i=2;i<=n;i++)

27         {

28             if(c>=bag[i].w)

29             {

30                 int temp = f[i-1][c-bag[i].w] + bag[i].value;

31                 f[i][c] = max(temp,f[i-1][c]);

32             }

33             else 

34                 f[i][c] = f[i-1][c];

35         }

36         cout<<f[n][c]<<endl;

37     }

38     return 0;

39 }

40         

 1 #include <iostream>

 2 #include <cstring>

 3 using namespace std;

 4 int w[1010];

 5 int value[1010];

 6 int f[1010][1010];

 7 int main()

 8 {

 9     int i,j,k,t,T;

10     cin>>T;

11     while(T--)

12     {

13         int n,c;

14         cin>>n>>c;

15         memset(w,0,sizeof(w));

16         memset(f,0,sizeof(f));

17         memset(value,0,sizeof(value));

18         for(i=1;i<=n;i++)

19             cin>>value[i];

20         for(i=1;i<=n;i++)

21             cin>>w[i];

22         for(i=1;i<=n;i++)

23             for(j=0;j<=c;j++)

24             if(j>=w[i])

25             {

26                 f[i][j] = max(f[i-1][j],f[i-1][j-w[i]] + value[i]);

27             }

28             else

29                 f[i][j] = f[i-1][j];

30                 /*

31                 //原来没加这一条语句

32                 1

33                 5 0

34                 2 4 1 5 1

35                 0 0 1 0 0

36                 答案应该是12

37                 却输出6 

38                 */ 

39             cout<<f[n][c]<<endl;

40     }

41     return 0;

42 }

43         

 

 1 //此时g++提交ac，c++wa，因为第二十九行的符号 

 2 #include <iostream>

 3 #include <cstring>

 4 using namespace std;

 5 int w[1010];

 6 int value[1010];

 7 int f[1010][1010];

 8 int main()

 9 {

10     int i,j,k,t,T;

11     cin>>T;

12     while(T--)

13     {

14         int n,c;

15         cin>>n>>c;

16         memset(w,0,sizeof(w));

17         memset(f,0,sizeof(f));

18         memset(value,0,sizeof(value));

19         for(i=1;i<=n;i++)

20             cin>>value[i];

21         for(i=1;i<=n;i++)

22             cin>>w[i];

23         for(i=1;i<=n;i++)

24             for(j=0;j<=c;j++)

25             {

26                 f[i][j] = f[i-1][j];

27                 if(j>=w[i])

28                 {

29                     f[i][j] >?= f[i-1][j-w[i]] + value[i];

30                 }

31             }

32             cout<<f[n][c]<<endl;

33     }

34     return 0;

35 }

36         

 

 1 //滚动数组优化空间和时间复杂度 ,时间上不太明显 

 2 #include <iostream>

 3 #include <cstring>

 4 using namespace std;

 5 int w[1010];

 6 int value[1010];

 7 int f[1010];

 8 int main()

 9 {

10     int i,j,k,t,T;

11     cin>>T;

12     while(T--)

13     {

14         int n,c;

15         cin>>n>>c;

16         memset(w,0,sizeof(w));

17         memset(f,0,sizeof(f));

18         memset(value,0,sizeof(value));

19         for(i=1;i<=n;i++)

20             cin>>value[i];

21         for(i=1;i<=n;i++)

22             cin>>w[i];

23         for(i=1;i<=n;i++)

24             for(j=c;j>=w[i];j--)

25             {

26                 f[j] >?= f[j-w[i]] + value[i];

27             }

28             cout<<f[c]<<endl;

29     }

30     return 0;

31 } 

32 //注意一条：不可用结构体存储重量和价值