hdu 1232 畅通工程

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8129    Accepted Submission(s): 4052

 本题非常简单就是锻炼学习并查集的，只要是能连在一起的就放在一个集合里面，最后看有多少个集合，然后用集合-1就是所要求得的结果；

代码：

 

  
    

   
     
   
      1 
   
     #include
   
     <
   
     stdio.h
   
     >
   
     

   
      2 
   
      
   
     int
   
      father[
   
     1005
   
     ],rank[
   
     1005
   
     ];

   
      3 
   
      
   
     int
   
      find(
   
     int
   
      x)

   
      4 
   
     {

   
      5 
   
      
   
     if
   
     (x
   
     !=
   
     father[x])

   
      6 
   
      father[x]
   
     =
   
     find(father[x]);

   
      7 
   
      
   
     return
   
      father[x];

   
      8 
   
     }

   
      9 
   
     
   
     void
   
      Union(
   
     int
   
      x,
   
     int
   
      y)

   
     10 
   
     {

   
     11 
   
      
   
     if
   
     (x
   
     ==
   
     y)

   
     12 
   
      
   
     return
   
      ;

   
     13 
   
      
   
     if
   
     (rank[x]
   
     >
   
     rank[y])

   
     14 
   
      father[y]
   
     =
   
     x;

   
     15 
   
      
   
     else
   
     

   
     16 
   
      {

   
     17 
   
      
   
     if
   
     (rank[x]
   
     ==
   
     rank[y])

   
     18 
   
      rank[y]
   
     ++
   
     ;

   
     19 
   
      father[x]
   
     =
   
     y;

   
     20 
   
      }

   
     21 
   
     }

   
     22 
   
     
   
     int
   
      main()

   
     23 
   
     {

   
     24 
   
      
   
     int
   
      n,m,i,k,f,ff,a,b,c[
   
     1005
   
     ],j;

   
     25 
   
      
   
     while
   
     (scanf(
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     n)
   
     !=
   
     EOF)

   
     26 
   
      {

   
     27 
   
      
   
     if
   
     (n
   
     ==
   
     0
   
     )

   
     28 
   
      
   
     break
   
     ;

   
     29 
   
      scanf(
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     m);

   
     30 
   
      
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     n;i
   
     ++
   
     )

   
     31 
   
      {

   
     32 
   
      father[i]
   
     =
   
     i;

   
     33 
   
      rank[i]
   
     =
   
     0
   
     ;

   
     34 
   
      }

   
     35 
   
      
   
     while
   
     (m
   
     --
   
     )

   
     36 
   
      {

   
     37 
   
      scanf(
   
     "
   
     %d%d
   
     "
   
     ,
   
     &
   
     a,
   
     &
   
     b);

   
     38 
   
      f
   
     =
   
     find(a);

   
     39 
   
      ff
   
     =
   
     find(b);

   
     40 
   
      Union(f,ff);

   
     41 
   
      }

   
     42 
   
      k
   
     =
   
     0
   
     ;

   
     43 
   
      
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     n;i
   
     ++
   
     )

   
     44 
   
      {

   
     45 
   
      f
   
     =
   
     find(i);

   
     46 
   
      
   
     for
   
     (j
   
     =
   
     1
   
     ;j
   
     <=
   
     k;j
   
     ++
   
     )

   
     47 
   
      
   
     if
   
     (f
   
     ==
   
     c[j])

   
     48 
   
      
   
     break
   
     ;

   
     49 
   
      
   
     if
   
     (j
   
     >
   
     k)

   
     50 
   
      c[
   
     ++
   
     k]
   
     =
   
     f;

   
     51 
   
      }

   
     52 
   
      printf(
   
     "
   
     %d\n
   
     "
   
     ,k
   
     -
   
     1
   
     );

   
     53 
   
      }

   
     54 
   
      
   
     return
   
      
   
     0
   
     ;

   
     55 
   
     }

   
     56 
   
     

   
     57 
   
     

   
     58