POJ 3041 Asteroids

解题思路：将行列分别对应二部图的两个set(行，列)->（left， right)，问题转化为最小顶点覆盖问题，进而转化为求解最大流问题

  
    

   
     
   
     #include 
   
     <
   
     iostream
   
     >
   
     

   
     using
   
      
   
     namespace
   
      std;

   
     #define
   
      Num 505
   
     


   
     bool
   
      edge[Num][Num];

   
     int
   
      father[Num];

   
     bool
   
      visit[Num];

   
     int
   
      N;


   
     int
   
      Search(
   
     int
   
      x)
{
 
   
     for
   
      (
   
     int
   
      i 
   
     =
   
      
   
     1
   
     ; i 
   
     <=
   
      N; i
   
     ++
   
     )
 {
 
   
     if
   
      (
   
     !
   
     visit[i] 
   
     &&
   
      edge[x][i])
 {
 visit[i] 
   
     =
   
      
   
     true
   
     ;
 
   
     if
   
      (father[i] 
   
     ==
   
      
   
     0
   
      
   
     ||
   
      Search(father[i]))
 {
 father[i] 
   
     =
   
      x;
 
   
     return
   
      
   
     1
   
     ;
 }
 }
 }
 
   
     return
   
      
   
     0
   
     ;
}


   
     int
   
      main()
{
 
   
     int
   
      K, R, C, t 
   
     =
   
      
   
     0
   
     ;
 scanf(
   
     "
   
     %d%d
   
     "
   
     , 
   
     &
   
     N, 
   
     &
   
     K);
 
   
     for
   
      (
   
     int
   
      i 
   
     =
   
      
   
     0
   
     ; i 
   
     <
   
      K; i
   
     ++
   
     )
 {
 scanf(
   
     "
   
     %d%d
   
     "
   
     , 
   
     &
   
     R, 
   
     &
   
     C);
 edge[R][C] 
   
     =
   
      
   
     true
   
     ;
 }
 
   
     for
   
      (
   
     int
   
      i 
   
     =
   
      
   
     1
   
     ; i 
   
     <=
   
      N; i
   
     ++
   
     )
 {
 memset(visit, 
   
     0
   
     , 
   
     sizeof
   
     (visit));
 t 
   
     +=
   
      Search(i);
 }
 printf(
   
     "
   
     %d\n
   
     "
   
     , t);
 
   
     return
   
      
   
     0
   
     ;
}