hdu 1102 Constructing Roads

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3230    Accepted Submission(s): 1082

      本题是简单的最小生成树问题，题意是说用最少的花费让他们全部连接起来，运用kruskals算法就能轻松搞定，因为连通两个分支，则这条线一定存在于最小生成树中

代码：

 

  
    

   
     
   
      1 
   
     #include
   
     <
   
     stdio.h
   
     >
   
     

   
      2 
   
     #include
   
     <
   
     queue
   
     >
   
     

   
      3 
   
      
   
     using
   
      
   
     namespace
   
      std;

   
      4 
   
      
   
     int
   
      father[
   
     105
   
     ],rank[
   
     105
   
     ];

   
      5 
   
     typedef 
   
     struct
   
      node

   
      6 
   
     {

   
      7 
   
      
   
     int
   
      x,y;

   
      8 
   
      
   
     int
   
      s;

   
      9 
   
      
   
     bool
   
      
   
     operator
   
     <
   
     (
   
     const
   
      node 
   
     &
   
     a)
   
     const
   
     

   
     10 
   
      {

   
     11 
   
      
   
     return
   
      a.s
   
     <
   
     s;

   
     12 
   
      }

   
     13 
   
     }NODE;

   
     14 
   
     
   
     int
   
      find(
   
     int
   
      x)

   
     15 
   
     {

   
     16 
   
      
   
     if
   
     (x
   
     !=
   
     father[x])

   
     17 
   
      {

   
     18 
   
      father[x]
   
     =
   
     find(father[x]);

   
     19 
   
      }

   
     20 
   
      
   
     return
   
      father[x];

   
     21 
   
     }

   
     22 
   
     
   
     void
   
      Union(
   
     int
   
      x,
   
     int
   
      y)

   
     23 
   
     {

   
     24 
   
      
   
     if
   
     (x
   
     ==
   
     y)

   
     25 
   
      
   
     return
   
      ;

   
     26 
   
      
   
     if
   
     (rank[x]
   
     >
   
     rank[y])

   
     27 
   
      {

   
     28 
   
      father[y]
   
     =
   
     x;

   
     29 
   
      }

   
     30 
   
      
   
     else
   
     

   
     31 
   
      {

   
     32 
   
      
   
     if
   
     (rank[x]
   
     ==
   
     rank[y])

   
     33 
   
      rank[y]
   
     ++
   
     ;

   
     34 
   
      father[x]
   
     =
   
     y;

   
     35 
   
      }

   
     36 
   
     }

   
     37 
   
     
   
     int
   
      main()

   
     38 
   
     {

   
     39 
   
      priority_queue
   
     <
   
     NODE
   
     >
   
      qu;

   
     40 
   
      NODE cur;

   
     41 
   
      
   
     int
   
      n,i,j,q,a,b,x,y,s;

   
     42 
   
      
   
     while
   
     (scanf(
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     n)
   
     !=
   
     EOF)

   
     43 
   
      {

   
     44 
   
      
   
     while
   
     (
   
     !
   
     qu.empty ())

   
     45 
   
      qu.pop();

   
     46 
   
      s
   
     =
   
     0
   
     ;

   
     47 
   
      
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     n;i
   
     ++
   
     )

   
     48 
   
      {

   
     49 
   
      father[i]
   
     =
   
     i;

   
     50 
   
      rank[i]
   
     =
   
     0
   
     ;

   
     51 
   
      
   
     for
   
     (j
   
     =
   
     1
   
     ;j
   
     <=
   
     n;j
   
     ++
   
     )

   
     52 
   
      {

   
     53 
   
      scanf(
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     a);

   
     54 
   
      
   
     if
   
     (i
   
     !=
   
     j)

   
     55 
   
      {

   
     56 
   
      cur.x
   
     =
   
     i;

   
     57 
   
      cur.y
   
     =
   
     j;

   
     58 
   
      cur.s
   
     =
   
     a;

   
     59 
   
      qu.push(cur);

   
     60 
   
      }

   
     61 
   
      }

   
     62 
   
      }

   
     63 
   
      scanf(
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     q);

   
     64 
   
      
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     q;i
   
     ++
   
     )

   
     65 
   
      {

   
     66 
   
      scanf(
   
     "
   
     %d%d
   
     "
   
     ,
   
     &
   
     a,
   
     &
   
     b);

   
     67 
   
      x
   
     =
   
     find(a);

   
     68 
   
      y
   
     =
   
     find(b);

   
     69 
   
      Union(x,y);

   
     70 
   
      }

   
     71 
   
      
   
     while
   
     (
   
     !
   
     qu.empty ())

   
     72 
   
      {

   
     73 
   
      cur
   
     =
   
     qu.top();

   
     74 
   
      qu.pop();

   
     75 
   
      x
   
     =
   
     cur.x;

   
     76 
   
      y
   
     =
   
     cur.y;

   
     77 
   
      
   
     if
   
     (find(x)
   
     !=
   
     find(y))

   
     78 
   
      {

   
     79 
   
      s
   
     +=
   
     cur.s;

   
     80 
   
      Union(find(x),find(y));

   
     81 
   
      }

   
     82 
   
      }

   
     83 
   
      printf(
   
     "
   
     %d\n
   
     "
   
     ,s);

   
     84 
   
      }

   
     85 
   
      
   
     return
   
      
   
     0
   
     ;

   
     86 
   
     }

   
     87 
   
     

   
     88 
   
     

   
     89 
   
      

   
     90 
   
      

   
     91 
   
      

   
     92