POJ 1789 Truck History

解题思路：prim算法求解最小生成树

  
    

   
     
   
     #include
   
     <
   
     iostream
   
     >
   
     

   
     using
   
      
   
     namespace
   
      std;


   
     #define
   
      MaxNum 1e8
   
     

inline 
   
     int
   
      distance(
   
     char
   
      
   
     *
   
     a, 
   
     char
   
      
   
     *
   
     b)
{
 
   
     int
   
      t 
   
     =
   
      
   
     0
   
     ;
 
   
     for
   
      (
   
     int
   
      i 
   
     =
   
      
   
     0
   
     ; i 
   
     <
   
      
   
     7
   
     ; i
   
     ++
   
     )
 
   
     if
   
      (a[i] 
   
     !=
   
      b[i])
 t
   
     ++
   
     ;
 
   
     return
   
      t;
}

   
     int
   
      main()
{
 
   
     int
   
      n, min, start, tempS, totalDist;
 
   
     char
   
      input[
   
     2001
   
     ][
   
     8
   
     ];
 
   
     int
   
      dist[
   
     2001
   
     ];
 
   
     bool
   
      visited[
   
     2001
   
     ];

 
   
     while
   
      (cin 
   
     >>
   
      n 
   
     &&
   
      n)
 {
 dist[
   
     0
   
     ] 
   
     =
   
      
   
     0
   
     ;
 start 
   
     =
   
      
   
     0
   
     ;

 fill(
   
     &
   
     dist[
   
     1
   
     ], 
   
     &
   
     dist[n], MaxNum);
 memset(visited, 
   
     0
   
     , 
   
     sizeof
   
     (
   
     bool
   
     ) 
   
     *
   
      
   
     2001
   
     );

 
   
     for
   
      (
   
     int
   
      i 
   
     =
   
      
   
     0
   
     ; i 
   
     <
   
      n; i
   
     ++
   
     )
 cin 
   
     >>
   
      input[i];

 
   
     while
   
      (start 
   
     !=
   
      
   
     -
   
     1
   
     )
 {
 visited[start] 
   
     =
   
      
   
     true
   
     ;
 tempS 
   
     =
   
      
   
     -
   
     1
   
     ;
 min 
   
     =
   
      MaxNum;

 
   
     for
   
      (
   
     int
   
      i 
   
     =
   
      
   
     0
   
     ; i 
   
     <
   
      n; i
   
     ++
   
     )
 {
 
   
     if
   
      (
   
     !
   
     visited[i])
 {
 
   
     if
   
      (dist[i] 
   
     !=
   
      
   
     1
   
     )
 {
 
   
     int
   
      l 
   
     =
   
      distance(input[start], input[i]);
 
   
     if
   
      (dist[i] 
   
     >
   
      l)
 dist[i] 
   
     =
   
      l;
 } 
 
   
     if
   
      (min 
   
     >
   
      dist[i])
 {
 min 
   
     =
   
      dist[i];
 tempS 
   
     =
   
      i;
 }
 }
 }
 start 
   
     =
   
      tempS;
 }
 totalDist 
   
     =
   
      
   
     0
   
     ;
 
   
     for
   
      (
   
     int
   
      i 
   
     =
   
      
   
     0
   
     ; i 
   
     <
   
      n; i
   
     ++
   
     )
 totalDist 
   
     +=
   
      dist[i];
 cout 
   
     <<
   
      
   
     "
   
     The highest possible quality is 1/
   
     "
   
      
   
     <<
   
      totalDist 
   
     <<
   
      
   
     "
   
     .
   
     "
   
      
   
     <<
   
      endl;
 }
 
   
     return
   
      
   
     0
   
     ;
}