hdu--1269--tarjan<有疑惑>

  1 //横插边：连接到已经出栈的节点的边；
  2 //后向边：连接到已在栈中节点的边；
  3 //树枝边: 在搜索树中的边。
  4 #include <iostream>
  5 #include <cstring>
  6 #include <stack>
  7 #include <algorithm>
  8 using namespace std;
  9 
 10 int cnt , n , sum , index;
 11 const int size = 10010;
 12 int No[size];
 13 int father[size];
 14 bool vis[size];
 15 struct graph
 16 {
 17     int to;
 18     int next;
 19 }edge[size*10];
 20 int head[size];
 21 stack<int>s;
 22 
 23 void init( )
 24 {
 25     memset( head , -1 , sizeof(head) );
 26     memset( vis , false , sizeof(vis) );
 27     memset( No , 0 , sizeof(No) );
 28 }
 29 
 30 void add( int st , int end )
 31 {
 32     while( !s.empty() )
 33         s.pop();
 34     edge[cnt].to = end;
 35     edge[cnt].next = head[st];
 36     head[st] = cnt++;
 37 }
 38 
 39 void tarjan( int u )
 40 {
 41     int temp , v;
 42     No[u] = father[u] = index++;
 43     vis[u] = true;
 44     s.push(u);
 45     for( int i = head[u] ; ~i ; i = edge[i].next )
 46     {
 47         v = edge[i].to;
 48         if( !No[v] )
 49         {
 50             tarjan(v);
 51             father[u] = min( father[u] , father[v] );
 52         }
 53         else if( vis[v] )
 54         {    
 55             father[u] = min( father[u] , No[v] );
 56         }
 57     }
 58     if( No[u] == father[u] )
 59     {
 60         sum ++;
 61         while(1)
 62         {
 63             temp = s.top();
 64             s.pop();
 65             vis[temp] = false;
 66             if( No[temp]==father[temp] )
 67                 break;
 68         }
 69     }
 70 }
 71 
 72 int main()
 73 {
 74     cin.sync_with_stdio(false);
 75     int m , x , y;
 76     while( cin >> n >> m &&(n||m) )
 77     {
 78         init();
 79          sum = cnt = 0;
 80          index = 1;
 81         while( m-- )
 82         {
 83             cin >> x >> y;
 84             add(x,y);
 85         }
 86         for( int i = 1 ; i<=n ; i++ )
 87         {
 88             if( !No[i] )
 89             {
 90                 tarjan(i);
 91             }
 92         }
 93         //for( int i = 1 ; i<= n ;i++ )
 94         //{
 95         //    cout << i << " " << No[i] << "  " << father[i] << endl;
 96         //}
 97         if( sum==1 )
 98             cout << "Yes" << endl;
 99         else
100             cout << "No" << endl;
101     }
102     return 0;
103 }
104 /*
105 6 8
106 1 2
107 2 4
108 4 6
109 5 6
110 3 4
111 3 5
112 4 1
113 1 3
114 标号 i  No[i]  father[i]
115   1       1        1
116   2       6        5
117   3       2        1
118   4       5        1
119   5       3        3
120   6       4        4
121  */