Bone Collector

Bone Collector

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 14   Accepted Submission(s) : 7

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

此题为典型的0-1背包问题，核心代码公式为 for(i=0;i<N;i++)
   {
    for(j=V;j>=volume[i];j--)
     f[j]=(f[j]>f[j-volume[i]]+value[i] ? f[j] : f[j-volume[i]]+value[i]);
   }

这题如果用多重背包解答的话存在问题，多重背包bag[value[i]][valume[j]]中value[i]和volume[j]的值相乘使得bag数组的空间开的过大，使得数组越界而出现RE错误。

所以不适合用多重背包来解

 

 1 #include<stdio.h>

 2 #include<string.h>

 3 int main()

 4 {

 5     int T,N,V,i,j;

 6     int f[1002],volume[1001],value[1001];

 7     while(scanf("%d",&T)==1)

 8     {

 9         while(T--)

10         {

11             scanf("%d %d",&N,&V);

12             for(i=0;i<N;i++)

13                 scanf("%d",&value[i]);

14             for(i=0;i<N;i++)

15                 scanf("%d",&volume[i]);

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

17             for(i=0;i<N;i++)

18             {

19                 for(j=V;j>=volume[i];j--)

20                     f[j]=(f[j]>f[j-volume[i]]+value[i] ? f[j] : f[j-volume[i]]+value[i]);

21             }

22             printf("%d\n",f[V]);

23         }

24     }

25     return 0;

26 }