Ubiquitous Religions--POJ 2524

1、解题思路：并查集。

2、注意事项：借用father[]、rank[]两个数组记录的小技巧。

3、实现方法：

  
    

   
     
   
      1 
   
     #include
   
     <
   
     iostream
   
     >
   
     

   
      2 
   
     #define
   
      M 50010
   
     

   
      3 
   
     using
   
      
   
     namespace
   
      std;

   
      4 
   
     

   
      5 
   
     int
   
      father[M],rank[M],ans;

   
      6 
   
     

   
      7 
   
     void
   
      Makeset(
   
     int
   
      x)

   
      8 
   
     {

   
      9 
   
      father[x]
   
     =
   
     x;

   
     10 
   
      rank[x]
   
     =
   
     0
   
     ;

   
     11 
   
     }

   
     12 
   
     

   
     13 
   
     int
   
      Findfather(
   
     int
   
      x)

   
     14 
   
     {

   
     15 
   
      
   
     if
   
     (x
   
     !=
   
     father[x])

   
     16 
   
      father[x]
   
     =
   
     Findfather(father[x]);

   
     17 
   
      
   
     return
   
      father[x];

   
     18 
   
     }

   
     19 
   
     

   
     20 
   
     void
   
      Union(
   
     int
   
      x,
   
     int
   
      y)

   
     21 
   
     {

   
     22 
   
      
   
     int
   
      a
   
     =
   
     Findfather(x);

   
     23 
   
      
   
     int
   
      b
   
     =
   
     Findfather(y);

   
     24 
   
      
   
     if
   
     (a
   
     !=
   
     b)

   
     25 
   
      {

   
     26 
   
      
   
     if
   
     (rank[a]
   
     >
   
     rank[b])

   
     27 
   
      father[b]
   
     =
   
     a;

   
     28 
   
      
   
     else
   
     

   
     29 
   
      {

   
     30 
   
      
   
     if
   
     (rank[a]
   
     ==
   
     rank[b])

   
     31 
   
      rank[b]
   
     ++
   
     ;

   
     32 
   
      father[a]
   
     =
   
     b;

   
     33 
   
      }

   
     34 
   
      ans
   
     --
   
     ;

   
     35 
   
      }

   
     36 
   
     }

   
     37 
   
     

   
     38 
   
     int
   
      main()

   
     39 
   
     {

   
     40 
   
      
   
     int
   
      i,j,n,m,tmp1,tmp2,cnt
   
     =
   
     0
   
     ;

   
     41 
   
      
   
     while
   
     (cin
   
     >>
   
     n
   
     >>
   
     m)

   
     42 
   
      {

   
     43 
   
      
   
     if
   
     (
   
     !
   
     n 
   
     &&
   
      
   
     !
   
     m)

   
     44 
   
      
   
     break
   
     ;

   
     45 
   
      
   
     for
   
     (j
   
     =
   
     1
   
     ;j
   
     <=
   
     n;j
   
     ++
   
     )

   
     46 
   
      Makeset(j);

   
     47 
   
      ans
   
     =
   
     n;

   
     48 
   
      
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     m;i
   
     ++
   
     )

   
     49 
   
      {

   
     50 
   
      scanf(
   
     "
   
     %d %d
   
     "
   
     ,
   
     &
   
     tmp1,
   
     &
   
     tmp2); 

   
     51 
   
      Union(tmp1,tmp2);

   
     52 
   
      }

   
     53 
   
      cout
   
     <<
   
     "
   
     Case 
   
     "
   
     <<++
   
     cnt
   
     <<
   
     "
   
     : 
   
     "
   
     <<
   
     ans
   
     <<
   
     endl;

   
     54 
   
      }

   
     55 
   
      
   
     return
   
      
   
     0
   
     ;

   
     56 
   
     }