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。

#include <stack>
#include <cstdio>
#include <cstring>
using namespace std;

const int N = 10001;
const int M = 100000;
int dfn[N], low[N], head[N];
bool inStack[N];
int n, m, e, cnt, dfs_clock;
stack<int> s;

struct Edge 
{
    int v, next;
} edge[M];

void init()
{
    e = cnt = dfs_clock = 0;
    memset( head, -1, sizeof(head) );
    memset( dfn, 0, sizeof(dfn) );
    memset( inStack, 0, sizeof(inStack) );
}

void addEdge( int u, int v )
{
    edge[e].v = v;
    edge[e].next = head[u];
    head[u] = e++;
}

void dfs( int u )
{
    dfn[u] = low[u] = ++dfs_clock;
    inStack[u] = true;
    s.push(u);
    for ( int i = head[u]; i != -1; i = edge[i].next )
    {
        int v = edge[i].v;
        if ( dfn[v] == 0 )
        {
            dfs(v);
            if ( low[v] < low[u] )
            {
                low[u] = low[v];
            }
        }
        else if ( inStack[v] )
        {
            if ( dfn[v] < low[u] )
            {
                low[u] = dfn[v];
            }
        }
    }
    if ( low[u] == dfn[u] )
    {
        cnt++;
        while ( 1 )
        {
            int tmp = s.top(); s.pop();
            inStack[tmp] = false;
            if ( tmp == u ) break;
        }
    }
}

void solve()
{
    for ( int i = 1; i <= n; i++ )
    {
        if ( !dfn[i] )
        {
            dfs(i);
        }
    }
}

int main ()
{
    while ( scanf("%d%d", &n, &m), n + m )
    {
        int u, v;
        init();
        while ( m-- )
        {
            scanf("%d%d", &u, &v);
            addEdge( u, v );
        }
        solve();
        printf("%s
", ( cnt == 1 ) ? "Yes" : "No");
    }
    return 0;
}