hoj2819

匈牙利算法。在二分图中每行对应一个节点，每列对应一个节点，每个1对应一条连接该1所在行和列节点的边。如果最大匹配数==n则可行，否则不行。

对于可行的情况，只需要每次交换两个列节点的编号，直到所有的匹配都是行号==列号的为止。

View Code

   
     

    
      #include 
    
      <
    
      iostream
    
      >
    
      
#include 
    
      <
    
      cstdlib
    
      >
    
      
#include 
    
      <
    
      cstring
    
      >
    
      
#include 
    
      <
    
      cstdio
    
      >
    
      
#include 
    
      <
    
      vector
    
      >
    
      

    
      using
    
       
    
      namespace
    
       std;


    
      #define
    
       maxn 105
    
      


    
      int
    
       n, ptn[maxn], ans[maxn][
    
      2
    
      ], tot;

    
      bool
    
       vis[maxn];
vector
    
      <
    
       vector
    
      <
    
      int
    
      >
    
       
    
      >
    
       G(maxn);


    
      void
    
       input()
{
 
    
      for
    
       (
    
      int
    
       i 
    
      =
    
       
    
      0
    
      ; i 
    
      <
    
       n; i
    
      ++
    
      )
 G[i].clear();
 
    
      for
    
       (
    
      int
    
       i 
    
      =
    
       
    
      0
    
      ; i 
    
      <
    
       n; i
    
      ++
    
      )
 
    
      for
    
       (
    
      int
    
       j 
    
      =
    
       
    
      0
    
      ; j 
    
      <
    
       n; j
    
      ++
    
      )
 {
 
    
      int
    
       a;
 scanf(
    
      "
    
      %d
    
      "
    
      , 
    
      &
    
      a);
 
    
      if
    
       (a 
    
      ==
    
       
    
      1
    
      )
 G[i].push_back(j);
 }
}


    
      bool
    
       find(
    
      int
    
       a)
{
 
    
      for
    
       (
    
      int
    
       i 
    
      =
    
       
    
      0
    
      ; i 
    
      <
    
       G[a].size(); i
    
      ++
    
      )
 
    
      if
    
       (
    
      !
    
      vis[G[a][i]])
 {
 
    
      if
    
       (ptn[G[a][i]] 
    
      ==
    
       
    
      -
    
      1
    
      )
 {
 ptn[G[a][i]] 
    
      =
    
       a;
 
    
      return
    
       
    
      true
    
      ;
 }
 vis[G[a][i]] 
    
      =
    
       
    
      true
    
      ;
 
    
      if
    
       (find(ptn[G[a][i]]))
 {
 ptn[G[a][i]] 
    
      =
    
       a;
 
    
      return
    
       
    
      true
    
      ;
 }
 }
 
    
      return
    
       
    
      false
    
      ;
}


    
      void
    
       work()
{
 tot 
    
      =
    
       
    
      0
    
      ;
 
    
      while
    
       (
    
      1
    
      )
 {
 
    
      int
    
       a 
    
      =
    
       
    
      -
    
      1
    
      ;
 
    
      for
    
       (
    
      int
    
       i 
    
      =
    
       
    
      0
    
      ; i 
    
      <
    
       n; i
    
      ++
    
      )
 
    
      if
    
       (ptn[i] 
    
      !=
    
       i)
 {
 a 
    
      =
    
       i;
 
    
      break
    
      ;
 }
 
    
      if
    
       (a 
    
      ==
    
       
    
      -
    
      1
    
      )
 
    
      return
    
      ;
 
    
      while
    
       (ptn[a] 
    
      !=
    
       a)
 {
 ans[tot][
    
      0
    
      ] 
    
      =
    
       a;
 ans[tot][
    
      1
    
      ] 
    
      =
    
       ptn[a];
 tot
    
      ++
    
      ;
 
    
      int
    
       x 
    
      =
    
       ptn[a];
 swap(ptn[a], ptn[x]);
 }
 }
}


    
      int
    
       main()
{
 
    
      //
    
      freopen("D:\\t.txt", "r", stdin);
    
      

    
       
    
      while
    
       (scanf(
    
      "
    
      %d
    
      "
    
      , 
    
      &
    
      n) 
    
      !=
    
       EOF 
    
      &&
    
       n 
    
      !=
    
       
    
      -
    
      1
    
      )
 {
 input();
 
    
      int
    
       ansi 
    
      =
    
       
    
      0
    
      ;
 memset(ptn, 
    
      -
    
      1
    
      , 
    
      sizeof
    
      (ptn));
 
    
      for
    
       (
    
      int
    
       i 
    
      =
    
       
    
      0
    
      ; i 
    
      <
    
       n; i
    
      ++
    
      )
 {
 memset(vis, 
    
      0
    
      , 
    
      sizeof
    
      (vis));
 
    
      if
    
       (find(i))
 ansi
    
      ++
    
      ;
 }
 
    
      if
    
       (ansi 
    
      <
    
       n)
 {
 printf(
    
      "
    
      -1\n
    
      "
    
      );
 
    
      continue
    
      ;
 }
 work();
 printf(
    
      "
    
      %d\n
    
      "
    
      , tot);
 
    
      for
    
       (
    
      int
    
       i 
    
      =
    
       
    
      0
    
      ; i 
    
      <
    
       tot; i
    
      ++
    
      )
 printf(
    
      "
    
      C %d %d\n
    
      "
    
      , ans[i][
    
      0
    
      ] 
    
      +
    
       
    
      1
    
      , ans[i][
    
      1
    
      ] 
    
      +
    
       
    
      1
    
      );
 }
 
    
      return
    
       
    
      0
    
      ;
}