hdu 1269 tarjan求强连通分量

tarjan求强连通分量的裸题复习，可当做模板。
 1 #include <stack>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 
 6 const int N = 10001;
 7 const int M = 100000;
 8 int dfn[N], low[N], head[N];
 9 bool inStack[N];
10 int n, m, e, cnt, dfs_clock;
11 stack<int> s;
12 
13 struct Edge 
14 {
15     int v, next;
16 } edge[M];
17 
18 void init()
19 {
20     e = cnt = dfs_clock = 0;
21     memset( head, -1, sizeof(head) );
22     memset( dfn, 0, sizeof(dfn) );
23     memset( inStack, 0, sizeof(inStack) );
24 }
25 
26 void addEdge( int u, int v )
27 {
28     edge[e].v = v;
29     edge[e].next = head[u];
30     head[u] = e++;
31 }
32 
33 void dfs( int u )
34 {
35     dfn[u] = low[u] = ++dfs_clock;
36     inStack[u] = true;
37     s.push(u);
38     for ( int i = head[u]; i != -1; i = edge[i].next )
39     {
40         int v = edge[i].v;
41         if ( dfn[v] == 0 )
42         {
43             dfs(v);
44             if ( low[v] < low[u] )
45             {
46                 low[u] = low[v];
47             }
48         }
49         else if ( inStack[v] )
50         {
51             if ( dfn[v] < low[u] )
52             {
53                 low[u] = dfn[v];
54             }
55         }
56     }
57     if ( low[u] == dfn[u] )
58     {
59         cnt++;
60         while ( 1 )
61         {
62             int tmp = s.top(); s.pop();
63             inStack[tmp] = false;
64             if ( tmp == u ) break;
65         }
66     }
67 }
68 
69 void solve()
70 {
71     for ( int i = 1; i <= n; i++ )
72     {
73         if ( !dfn[i] )
74         {
75             dfs(i);
76         }
77     }
78 }
79 
80 int main ()
81 {
82     while ( scanf("%d%d", &n, &m), n + m )
83     {
84         int u, v;
85         init();
86         while ( m-- )
87         {
88             scanf("%d%d", &u, &v);
89             addEdge( u, v );
90         }
91         solve();
92         printf("%s
", ( cnt == 1 ) ? "Yes" : "No");
93     }
94     return 0;
95 }