Bone Collector

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

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

 

Sample Output

14

 

#include

using namespace std;

int p[1001][1001];

int max(int a,int b){

if(a>=b)

return a;

else

return b;

}

int main(){

int a[1000],v[1000];

int t,n,c;

scanf("%d",&t);

while(t--){

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

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

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

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

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

memset(p,0,sizeof(p));



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

for(int j=0;j<=c;j++){

if(j>=v[i])

p[i][j]=max(p[i-1][j],p[i-1][j-v[i]]+a[i]);

else

p[i][j]=p[i-1][j];

}



/* for(int i=0;i
for(int j=c;j>=v[i];j--){

p[j]=max(p[j],p[j-v[i]]+a[i]);

}

*/

printf("%d ",p[n][c]);



}

return 0;

}

以上两种方法都可以接受，但注释的方法更高效

解释，可参考http://blog.sina.com.cn/s/blog_7e5541250100rtv5.html