强连通分量--tarjan算法

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int m,n,len = 0,stc[50050];
int ri = 0,tot = 0,ans = 0;
int num[50005];
struct node{
    int l,r,nxt;
}a[50050];
int low[50050],lst[50050];
int dn[50080];
int chu[50050];
bool vis[50050];
int add(int x,int y)
{
    a[++len].l = x;
    a[len].r = y;
    a[len].nxt = lst[x];
    lst[x] = len;
    chu[x]++;
}
void dfs(int x)
{
    dn[x] = low[x] = ++tot;
    stc[++ri] = x;
    vis[x] = 1;
    for(int k = lst[x];k;k = a[k].nxt)
    {
        int y = a[k].r;
        if(!dn[y])
        {
            dfs(y);
            low[x] = min(low[x],low[y]);
        }
        else if(vis[y])
        {
            low[x] = min(low[x],dn[y]);
        }
    }
    if(low[x] == dn[x])
    {
        ans ++;
        int v;
        do
        {
            num[ans]++;
            vis[stc[ri]] = 0;
            v = stc[ri--];
        }
        while(x != v);
    }
}
int main()
{
    memset(vis,0,sizeof(vis));
    scanf("%d%d",&n,&m);
    int x,y;
    for(int i = 1;i <= m;i++)
    {
        scanf("%d%d",&x,&y);
        add(x,y);
    }
    for(int i = 1;i <= n;i++)
        if(dn[i] == 0)
        {
            dfs(i);
        }
//    cout<<ans<<endl;
    tot = 0;
    for(int i = 1;i <= ans;i++)
    {
        if(num[i] > 1)
        {
            tot++;
        }
    }
    printf("%d\n",tot);
    return 0;
}