POJ 3020 Antenna Placement

解题思路:求解最大独立集，转而求解最小顶点覆盖->最大流问题

  
    

   
     
   
     #include 
   
     <
   
     iostream
   
     >
   
     

   
     using
   
      
   
     namespace
   
      std;


   
     int
   
      t;

   
     int
   
      father[
   
     405
   
     ];

   
     bool
   
      visit[
   
     405
   
     ];

   
     struct
   
     
{
 
   
     int
   
      x,y;
}point[
   
     405
   
     ];


   
     int
   
      Search(
   
     int
   
      x)
{
 
   
     for
   
      (
   
     int
   
      i 
   
     =
   
      
   
     1
   
     ; i 
   
     <
   
      t; i
   
     ++
   
     )
 {
 
   
     if
   
      (
   
     !
   
     visit[i] 
   
     &&
   
      (abspoint[x].x 
   
     -
   
      point[i].x) 
   
     +
   
      abs(point[x].y 
   
     -
   
      point[i].y)) 
   
     ==
   
      
   
     1
   
     )
 {
 visit[i] 
   
     =
   
      
   
     true
   
     ;
 
   
     if
   
      (father[i] 
   
     ==
   
      
   
     0
   
      
   
     ||
   
      Search(father[i]))
 {
 father[i] 
   
     =
   
      x;
 
   
     return
   
      
   
     1
   
     ;
 }
 }
 }
 
   
     return
   
      
   
     0
   
     ;
}


   
     int
   
      main()
{
 
   
     int
   
      k, n, m, u, v, s, row, col;
 
   
     char
   
      ch;
 
   
     const
   
      
   
     int
   
      dir[
   
     4
   
     ][
   
     2
   
     ] 
   
     =
   
      {{
   
     -
   
     1
   
     , 
   
     0
   
     }, {
   
     1
   
     , 
   
     0
   
     }, {
   
     0
   
     , 
   
     -
   
     1
   
     }, {
   
     0
   
     , 
   
     1
   
     }};
 cin 
   
     >>
   
      k;
 
   
     while
   
      (k
   
     --
   
     )
 {
 memset(father, 
   
     0
   
     , 
   
     sizeof
   
     (father));
 t 
   
     =
   
      
   
     1
   
     ; s 
   
     =
   
      
   
     0
   
     ;
 cin 
   
     >>
   
      n 
   
     >>
   
      m;
 
   
     for
   
      (
   
     int
   
      i 
   
     =
   
      
   
     0
   
     ; i 
   
     <
   
      n; i
   
     ++
   
     )
 
   
     for
   
      (
   
     int
   
      j 
   
     =
   
      
   
     0
   
     ; j 
   
     <
   
      m; j
   
     ++
   
     )
 {
 cin 
   
     >>
   
      ch;
 
   
     if
   
      (ch 
   
     ==
   
      
   
     '
   
     *
   
     '
   
     )
 {
 point[t].y 
   
     =
   
      i;
 point[t
   
     ++
   
     ].x 
   
     =
   
      j;
 }
 }
 
   
     for
   
      (
   
     int
   
      i 
   
     =
   
      
   
     1
   
     ; i 
   
     <
   
      t; i
   
     ++
   
     )
 {
 memset(visit, 
   
     0
   
     , 
   
     sizeof
   
     (visit));
 s 
   
     +=
   
      Search(i);
 }
 t 
   
     =
   
      t 
   
     -
   
      s 
   
     /
   
      
   
     2
   
      
   
     -
   
      
   
     1
   
     ;
 cout 
   
     <<
   
      t 
   
     <<
   
      endl;;
 }
 
   
     return
   
      
   
     0
   
     ;
}