Hdu4496 d-city [concurrent search]

Topic link ：

​ ​http://acm.hdu.edu.cn/showproblem.php?pid=4496​​

The main idea of the topic ：

Give a picture , Delete one by one according to the order of the given edges . Ask the number of connected blocks of the graph after deleting each edge .

Ideas ：

Reverse and search the set to find the number of connected blocks . Suppose at the beginning, all points are disconnected . Reverse the order from the given edge , That is, from the end

An edge begins to be added . If the newly added edge connects two connected components , Then the number of connected blocks will be reduced by one , Otherwise it doesn't change . take

Save the number of connected blocks after each edge addition . Finally output .

AC Code ：

Log in and copy

      
      

       
       #include<iostream>
       
       

       
       #include<algorithm>
       
       

       
       #include<cstdio>
       
       

       
       #include<cstring>
       
       

       
       using 
       
       namespace 
       
       std;
       
       

       
       

       
       struct 
       
       EdgeNode
       
       
{
       
       
    
       
       int 
       
       u;
       
       
    
       
       int 
       
       v;
       
       
}
       
       Edges[
       
       1000010];
       
       

       
       int 
       
       Father[
       
       100010],
       
       Ans[
       
       100010];
       
       

       
       

       
       int 
       
       Find(
       
       int 
       
       x)
       
       
{
       
       
    
       
       if(
       
       Father[
       
       x] 
       
       != 
       
       x)
       
       
        
       
       Father[
       
       x] 
       
       = 
       
       Find(
       
       Father[
       
       x]);
       
       
    
       
       return 
       
       Father[
       
       x];
       
       
}
       
       

       
       

       
       int 
       
       main()
       
       
{
       
       
    
       
       int 
       
       N,
       
       M;
       
       
    
       
       while(
       
       ~scanf(
       
       "%d%d",
       
       &
       
       N,
       
       &
       
       M))
       
       
    {
       
       
        
       
       for(
       
       int 
       
       i 
       
       = 
       
       0; 
       
       i 
       
       < 
       
       M; 
       
       ++
       
       i)
       
       
            
       
       scanf(
       
       "%d%d",
       
       &
       
       Edges[
       
       i].
       
       u,
       
       &
       
       Edges[
       
       i].
       
       v);
       
       
        
       
       for(
       
       int 
       
       i 
       
       = 
       
       0; 
       
       i 
       
       < 
       
       N; 
       
       ++
       
       i)
       
       
            
       
       Father[
       
       i] 
       
       = 
       
       i;
       
       
        
       
       int 
       
       j 
       
       = 
       
       0;
       
       
        
       
       int 
       
       Num 
       
       = 
       
       N;
       
       
        
       
       for(
       
       int 
       
       i 
       
       = 
       
       M
       
       -
       
       1; 
       
       i 
       
       >= 
       
       0; 
       
       --
       
       i)
       
       
        {
       
       
            
       
       int 
       
       u 
       
       = 
       
       Edges[
       
       i].
       
       u;
       
       
            
       
       int 
       
       v 
       
       = 
       
       Edges[
       
       i].
       
       v;
       
       
            
       
       u 
       
       = 
       
       Find(
       
       u);
       
       
            
       
       v 
       
       = 
       
       Find(
       
       v);
       
       
            
       
       if(
       
       u 
       
       != 
       
       v)
       
       
            {
       
       
                
       
       Father[
       
       u] 
       
       = 
       
       v;
       
       
                
       
       Ans[
       
       j
       
       ++] 
       
       = 
       
       Num;
       
       
                
       
       Num
       
       --;
       
       
            }
       
       
            
       
       else
       
       
                
       
       Ans[
       
       j
       
       ++] 
       
       = 
       
       Num;
       
       
        }
       
       
        
       
       for(
       
       int 
       
       i 
       
       = 
       
       j
       
       -
       
       1; 
       
       i 
       
       >= 
       
       0; 
       
       --
       
       i)
       
       
            
       
       printf(
       
       "%d\n",
       
       Ans[
       
       i]);
       
       
    }
       
       

       
       

       
       
    
       
       return 
       
       0;
       
       
}
      
      
      
      

       
       
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