hdu1455 Sticks (search+pruning+pruning+.....+pruning)

Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10325    Accepted Submission(s): 3091

Problem Description

George took sticks of the same length and cut them randomly until all parts became at most 50 units long. Now he wants to return sticks to the original state, but he forgot how many sticks he had originally and how long they were originally. Please help him and design a program which computes the smallest possible original length of those sticks. All lengths expressed in units are integers greater than zero. 

 

Input

The input contains blocks of 2 lines. The first line contains the number of sticks parts after cutting, there are at most 64 sticks. The second line contains the lengths of those parts separated by the space. The last line of the file contains zero.

 

Output

The output file contains the smallest possible length of original sticks, one per line. 

 

Sample Input

9 5 2 1 5 2 1 5 2 1 4 1 2 3 4 0

 

Sample Output

6 5

 

Source

​ ​ACM暑期集训队练习赛（七）​​

 

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​ ​Note​​

I feel like I'm going to die  参考了别人的代码  understood overnight

自己敲了下.在hdu过了  可是在​ ​nyoj293​​死活TLE..

Obviously the same code as others.last order  变量名 什么都改  还是不行..我选择死亡  nyojI don't brush it anymore

对于这道题而言 The importance of pruning in search is vividly reflected.

剪枝1：The length of the stick The longer the binding, the greater the force,less flexibility（The fewer lengths that can be combined）,So first sort the length of the small sticks

剪枝2：The sum of the lengths of all the sticks must divide the length of the original stick

剪枝3：If we have used wooden sticks4 5 6Unable to complete stitching,下次再出现5 4 6肯定是不可能的 So avoid this when searching

剪枝4：If the result of the current search is correct,for each small stick 都要使用,如果发现没有使用  直接退出

剪枝5：If the length of the current stick is equal to the length needed to splice the stick  ,And the stick was not used,那么直接退出.（Even find a few combinations smaller than him for this stick length 也是无用）

剪枝6：If the current length of the stick has already been present and has not been used 那么直接跳过

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       #include <stdio.h>
       
       

       
       #include <math.h>
       
       

       
       #include <string.h>
       
       

       
       #include <algorithm>
       
       

       
       using 
       
       namespace 
       
       std;
       
       

       
       int 
       
       a[
       
       100];
       
       

       
       int 
       
       n;
       
       

       
       bool 
       
       vis[
       
       100];
       
       

       
       bool 
       
       cmp(
       
       int 
       
       a, 
       
       int 
       
       b)
       
       
{
       
       
  
       
       return 
       
       a 
       
       > 
       
       b;
       
       
}
       
       

       
       bool 
       
       dfs(
       
       int 
       
       s, 
       
       int 
       
       c, 
       
       int 
       
       l, 
       
       int 
       
       index, 
       
       int 
       
       sum)
       
       
{
       
       
  
       
       if(
       
       sum 
       
       == 
       
       l)
       
       
  {
       
       
    
       
       index
       
       ++;
       
       
    
       
       if(
       
       index 
       
       == 
       
       c) 
       
       
    
       
       return 
       
       true;
       
       
    
       
       s 
       
       = 
       
       0;
       
       
    
       
       sum 
       
       = 
       
       0;
       
       
  }
       
       
  
       
       for(
       
       int 
       
       i 
       
       = 
       
       s; 
       
       i 
       
       < 
       
       n; 
       
       i
       
       ++)
       
       //剪枝3 
       
       
  {
       
       
    
       
       if(
       
       !
       
       vis[
       
       i] 
       
       && 
       
       sum 
       
       + 
       
       a[
       
       i] 
       
       <= 
       
       l)
       
       
    {
       
       
      
       
       vis[
       
       i] 
       
       = 
       
       true;
       
       
      
       
       if(
       
       dfs(
       
       s 
       
       + 
       
       1, 
       
       c, 
       
       l, 
       
       index, 
       
       sum 
       
       + 
       
       a[
       
       i])) 
       
       return 
       
       true; 
       
       
      
       
       vis[
       
       i] 
       
       = 
       
       false;
       
       
      
       
       if(
       
       sum 
       
       == 
       
       0) 
       
       return 
       
       false;
       
       //剪枝4：If this stick can't find a match 直接结束 
       
       
      
       
       if(
       
       l 
       
       - 
       
       sum 
       
       == 
       
       a[
       
       i]) 
       
       return 
       
       false;
       
       //剪枝5：If the length of the current stick is equal to the desired length 直接结束 
       
       
      
       
       while(
       
       a[
       
       i] 
       
       == 
       
       a[
       
       i 
       
       + 
       
       1]) 
       
       i
       
       ++;
       
       //剪枝6：去除重复的 
       
       
    }
       
       
  }
       
       
  
       
       return 
       
       false;
       
       
}
       
       

       
       int 
       
       main()
       
       
{
       
       
  
       
       while(
       
       scanf(
       
       "%d",
       
       &
       
       n)
       
       &&
       
       n)
       
       
  {
       
       
    
       
       int 
       
       sum
       
       =
       
       0;
       
       
    
       
       for(
       
       int 
       
       i 
       
       = 
       
       0; 
       
       i 
       
       < 
       
       n; 
       
       i
       
       ++)
       
       
    {
       
       
      
       
       scanf(
       
       "%d", 
       
       &
       
       a[
       
       i]);
       
       
      
       
       sum 
       
       += 
       
       a[
       
       i];
       
       
    }
       
       
    
       
       sort(
       
       a, 
       
       a 
       
       + 
       
       n, 
       
       cmp);
       
       //剪枝1 
       
       
    
       
       for(
       
       int 
       
       i 
       
       = 
       
       a[
       
       0]; 
       
       i 
       
       <= 
       
       sum; 
       
       i
       
       ++)
       
       
    {
       
       
      
       
       if(
       
       sum 
       
       % 
       
       i 
       
       == 
       
       0)
       
       //剪枝2 
       
       
      {
       
       
        
       
       memset(
       
       vis, 
       
       false, 
       
       sizeof(
       
       vis));
       
       
        
       
       if(
       
       dfs(
       
       0, 
       
       sum 
       
       / 
       
       i, 
       
       i, 
       
       0, 
       
       0))
       
       
        {
       
       
          
       
       printf(
       
       "%d\n", 
       
       i);
       
       
          
       
       break;
       
       
        }
       
       
      }
       
       
    }
       
       
  }
       
       
  
       
       return 
       
       0;
       
       
}
      
      
      
      

       
       
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