Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 29109    Accepted Submission(s): 11898

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

 

这是 01 背包问题 01 背包问题用到了递推的思想。而且用到了分治

将大问题转化为可以easy解决的，小问题。从小问题来得到答案的思想

大致思路例如以下：

c[i][v]表示前i-1件物品恰放入一个质量为m的背包能够取到最大价值。若仅仅考虑第i件物品的话，那么有两种情况。就是放，与不放，要是放的话，那么问题就转化为“将前n-1件物品放入剩下的容量为m-w[i]的背包中，此时获得的最大价值就是c[i-1][m-w[i]]再加上最后放进去的第i件的物品所获得价值p[i];要是不放的话。问题就变成：”将前n-1件物品放入容量为j的背包中。此时获得的最大价值就是c[i-1][m];然后题上一般有物品的数量，及对应的价值，通过两个for循环就好了。最后的最大价值就是c[n][m];

代码例如以下：

#include<stdio.h>

#include<string.h>

int dp[1010][1010],val[1010],vol[1010];

int main()

{

 int n,i,j,v,t;

 scanf("%d",&t);

 while(t--)

 {

  scanf("%d%d",&n,&v);

  for(i=1;i<=n;i++)

  scanf("%d",&val[i]);

  for(j=1;j<=n;j++)

  scanf("%d",&vol[j]);

  memset(dp,0,sizeof(dp));

  for(i=1;i<=n;i++)

  for(j=0;j<=v;j++)//题上有个陷阱。就是将没有体积可是有价值的 骨头考虑了进去

  {

   if(vol[i]<=j&&dp[i-1][j]<dp[i-1][j-vol[i]]+val[i])

   dp[i][j]=dp[i-1][j-vol[i]]+val[i];//放进去的情况

   else

   dp[i][j]=dp[i-1][j];//不放的情况

  }

  printf("%d\n",dp[n][v]);

 }

 return 0;

}