这是一道强连通分量板子题。

我们只用输出点数大于1的强连通分量的个数！

#include
    
     #include
     
      #include
      
       #define N 10010#define M 50010using namespace std;int n,m,head[N],dfn[N],low[N],cnt=1,t,sum,bel[N],num[N],out[N],ans;bool instk[N];stack 
       
         stk;struct hhh{    int to,next;}edge[M];int read(){    int ans=0,fu=1;    char j=getchar();    for (;(j<'0' || j>'9') && j!='-';j=getchar()) ;    if (j=='-') fu=-1,j=getchar();    for (;j>='0' && j<='9';j=getchar()) ans*=10,ans+=j-'0';    return ans*fu;}void add(int u,int v){    edge[cnt].to=v;    edge[cnt].next=head[u];    head[u]=cnt++;}void Tarjan(int x){    dfn[x]=low[x]=++t;    stk.push(x);    instk[x]=1;    int v;    for (int i=head[x];i;i=edge[i].next)    {    v=edge[i].to;    if (!dfn[v])    {        Tarjan(v);        low[x]=min(low[x],low[v]);    }    else if (instk[v]) low[x]=min(low[x],dfn[v]);    }    if (dfn[x]==low[x])    {    sum++;    do    {        v=stk.top();        stk.pop();        instk[v]=0;        bel[v]=sum;        num[sum]++;    }while(v!=x);    }}int main(){    n=read();    m=read();    for (int i=1,a,b;i<=m;i++)    {    a=read();    b=read();    add(a,b);    }    for (int i=1;i<=n;i++)    if (!dfn[i]) Tarjan(i);    for (int i=1;i<=sum;i++)    if (num[i]>1) ans++;    printf("%d",ans);    return 0;}