PKU 2239 Selecting Courses(二分图最大匹配)

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       题意大概是学校有多门课。没门课可能会在一个星期的每一天i(
    
      1
    
      <=
    
      i
    
      <=
    
      7
    
      )的第j(
    
      1
    
      <=
    
      j
    
      <=
    
      12
    
      )节课进行。求一个选课方案使得能选的课最多。
构造二分图,M
    
      =
    
      {课程}，N
    
      =
    
      {课程举行的时间},然后构造M到N得二分图。这里将时间(i,j)映射为一个整数(i
    
      -
    
      1
    
      )
    
      *
    
      12
    
      +
    
      j用于构图。然后用匈牙利算法进行匹配就行了。

#include 
    
      <
    
      iostream
    
      >
    
      
#include 
    
      <
    
      cstring
    
      >
    
      


    
      using
    
       
    
      namespace
    
       std;

    
      int
    
       map[
    
      305
    
      ][
    
      86
    
      ],match[
    
      86
    
      ],course[
    
      10
    
      ][
    
      15
    
       ];

    
      bool
    
       vis[
    
      86
    
      ];

    
      int
    
       n,t;


    
      bool
    
       findPath(
    
      int
    
       k)
{
 
    
      for
    
      (
    
      int
    
       i 
    
      =
    
       
    
      1
    
      ;i 
    
      <=
    
       
    
      84
    
      ;
    
      ++
    
      i)
 {
 
    
      if
    
      (
    
      !
    
      vis[i] 
    
      &&
    
       map[k][i])
 {
 vis[i] 
    
      =
    
       
    
      true
    
      ;

 
    
      if
    
      (
    
      !
    
      match[i] 
    
      ||
    
       findPath(match[i]))
 {
 match[i] 
    
      =
    
       k; 
    
      return
    
       
    
      true
    
      ;
 }
 }
 }

 
    
      return
    
       
    
      false
    
      ;
}


    
      int
    
       main()
{
 
    
      int
    
       i,j,time1,time2;

 
    
      for
    
      (i 
    
      =
    
       
    
      1
    
      ;i 
    
      <=
    
       
    
      7
    
      ;
    
      ++
    
      i)
 {
 
    
      for
    
      (j 
    
      =
    
       
    
      1
    
      ;j 
    
      <=
    
       
    
      12
    
      ;
    
      ++
    
      j) course[i][j] 
    
      =
    
       (i
    
      -
    
      1
    
      )
    
      *
    
      12
    
       
    
      +
    
       j;
 }

 
    
      while
    
      (scanf(
    
      "
    
      %d
    
      "
    
      ,
    
      &
    
      n) 
    
      !=
    
       EOF)
 {
 memset(map,
    
      0
    
      ,
    
      sizeof
    
      (map));
 
    
      for
    
      (i 
    
      =
    
       
    
      1
    
      ;i 
    
      <=
    
       n;
    
      ++
    
      i)
 {
 scanf(
    
      "
    
      %d
    
      "
    
      ,
    
      &
    
      t);

 
    
      while
    
      (t
    
      --
    
      )
 {
 scanf(
    
      "
    
      %d %d
    
      "
    
      ,
    
      &
    
      time1,
    
      &
    
      time2);
 map[i][course[time1][time2]] 
    
      =
    
       
    
      1
    
      ;
 }
 }

 
    
      int
    
       ans 
    
      =
    
       
    
      0
    
      ; memset(match,
    
      0
    
      ,
    
      sizeof
    
      (match));

 
    
      for
    
      (i 
    
      =
    
       
    
      1
    
      ;i 
    
      <=
    
       n;
    
      ++
    
      i)
 {
 memset(vis,
    
      false
    
      ,
    
      sizeof
    
      (vis));
 
    
      if
    
      (findPath(i)) ans
    
      ++
    
      ;
 }
 printf(
    
      "
    
      %d\n
    
      "
    
      ,ans);
 }

 
    
      return
    
       
    
      0
    
      ;
}