Popular Cows--POJ 2186

1、题目类型：图论、强连通分量、Tarjan算法。

2、解题思路：（1）建立图的邻接表；（2）Trajan算法求解并记录强连通分量；（3）判断强连通分量的入度，当存在多个出度为0时，输出0；当只存在唯一出度为0的强连通分量，则输出其内部节点的个数。

3、注意事项：图用邻接表表示，矩阵表示MLE；注意M=0的情况。

4、实现方法:

  
    

   
     
   
     #include
   
     <
   
     iostream
   
     >
   
     
#include
   
     <
   
     stack
   
     >
   
     

   
     using
   
      
   
     namespace
   
      std;


   
     struct
   
      TNode
{
 
   
     int
   
      data;
 TNode 
   
     *
   
     next;
};


   
     int
   
      n,m,index,cnt,bcnt;

   
     int
   
      dfn[
   
     10010
   
     ],low[
   
     10010
   
     ];

   
     int
   
      a[
   
     50010
   
     ],b[
   
     50010
   
     ],flag[
   
     10010
   
     ];
TNode 
   
     *
   
     map[
   
     10010
   
     ],data[
   
     500100
   
     ];

   
     bool
   
      vis[
   
     10010
   
     ];
stack
   
     <
   
     int
   
     >
   
      S;


   
     void
   
      Tarjan(
   
     int
   
      v)
{
 TNode 
   
     *
   
     tmp;
 dfn[v]
   
     =
   
     low[v]
   
     =++
   
     cnt;
 vis[v]
   
     =
   
     true
   
     ;
 S.push(v);
 
   
     for
   
     (tmp
   
     =
   
     map[v];tmp
   
     !=
   
     NULL;tmp
   
     =
   
     tmp
   
     ->
   
     next)
 {
 
   
     int
   
      i
   
     =
   
     tmp
   
     ->
   
     data;
 
   
     if
   
     (
   
     !
   
     dfn[i])
 {
 Tarjan(i);
 
   
     if
   
     (low[v]
   
     >
   
     low[i])
 low[v]
   
     =
   
     low[i];
 }
 
   
     else
   
      
   
     if
   
     (vis[i]
   
     &&
   
     low[v]
   
     >
   
     dfn[i])
 {
 low[v]
   
     =
   
     dfn[i];
 }
 } 
 
   
     if
   
     (low[v]
   
     ==
   
     dfn[v])
 {
 bcnt
   
     ++
   
     ;
 
   
     int
   
      j;
 
   
     do
   
     
 { 
 j
   
     =
   
     S.top();
 S.pop();
 
   
     //
   
     用于标示点j在第bcnt个强连通分量里面 
   
     

   
      flag[j]
   
     =
   
     bcnt; 
 }
   
     while
   
     (j
   
     !=
   
     v);
 }
}


   
     int
   
      main()
{
 
   
     int
   
      i,ans,pos;
 TNode 
   
     *
   
     P,
   
     *
   
     pre;
 cin
   
     >>
   
     n
   
     >>
   
     m;
 
   
     if
   
     (m
   
     ==
   
     0
   
     )
 {
 cout
   
     <<
   
     0
   
     <<
   
     endl;
 
   
     return
   
      
   
     0
   
     ;
 }
 
   
     //
   
     建立邻接表
   
     

   
      
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     m;i
   
     ++
   
     )
 {
 cin
   
     >>
   
     a[i]
   
     >>
   
     b[i];
 P
   
     =&
   
     data[index
   
     ++
   
     ];
 P
   
     ->
   
     data
   
     =
   
     b[i];
 
   
     if
   
     (map[a[i]]
   
     ==
   
     NULL)
 {
 map[a[i]]
   
     =
   
     P;
 }
 
   
     else
   
     
 {
 pre
   
     =
   
     map[a[i]]
   
     ->
   
     next;
 P
   
     ->
   
     next
   
     =
   
     pre;
 map[a[i]]
   
     ->
   
     next
   
     =
   
     P;
 }
 }
 
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     n;i
   
     ++
   
     )
 {
 
   
     if
   
     (
   
     !
   
     dfn[i])
 Tarjan(i);
 }
 
 memset(vis,
   
     0
   
     ,
   
     sizeof
   
     (vis));
 
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     m;i
   
     ++
   
     )
 {
 
   
     if
   
     (flag[a[i]]
   
     !=
   
     flag[b[i]])
 vis[flag[a[i]]]
   
     =
   
     1
   
     ;
 }
 cnt
   
     =
   
     0
   
     ;
 
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     bcnt;i
   
     ++
   
     )
 {
 
   
     if
   
     (
   
     !
   
     vis[i])
 {
 cnt
   
     ++
   
     ;
 pos
   
     =
   
     i;
 }
 }
 
   
     if
   
     (cnt
   
     >
   
     1
   
     )
 {
 cout
   
     <<
   
     0
   
     <<
   
     endl;
 }
 
   
     else
   
     
 {
 ans
   
     =
   
     0
   
     ;
 
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <=
   
     n;i
   
     ++
   
     )
 
   
     if
   
     (flag[i]
   
     ==
   
     pos)
 ans
   
     ++
   
     ;
 cout
   
     <<
   
     ans
   
     <<
   
     endl;
 }
 
   
     return
   
      
   
     0
   
     ;
}