首先做一遍tarjan

然后枚举每个边判断是否在同一个连通块内

在新图上跑一边入度

若入度为0，ans累计

输出即可

#include
     
      using namespace std;const int N=2e7+7;int m,n,k,p,idx,cnt,num,ans;struct node{    int nx,to;} e[N];bool vis[N];stack
      
        s;int head[N],dfn[N],low[N],sum[N],color[N],in[N];void add_edge(int a,int b){    cnt++;e[cnt].to=b;e[cnt].nx=head[a];head[a]=cnt;}void paint(int x){    s.pop();    color[x]=num;    sum[num]++;    vis[x]=false;}void tarjan(int x){    dfn[x]=low[x]=++idx;    s.push(x);vis[x]=true;    for (int i=head[x];i;i=e[i].nx)    {        int y=e[i].to;        if (!dfn[y])        {            tarjan(y);            low[x]=min(low[x],low[y]);        }        else if (vis[y]) low[x]=min(low[x],dfn[y]);    }    if (dfn[x]==low[x])    {        num++;        while (s.top()!=x)        {            int t=s.top();            paint(t);        }        paint(x);    }}int main(){    scanf("%d%d",&n,&m);    for (int i=1;i<=m;i++)    {        int x,y;        scanf("%d%d",&x,&y);        add_edge(x,y);    }    for (int i=1;i<=n;i++)    if (!dfn[i]) tarjan(i);    for (int i=1;i<=n;i++)    {     for (int j=head[i];j;j=e[j].nx)     {         int y=e[j].to;         if (color[i]==color[y]) continue;         in[color[y]]++;     }    }    for (int i=1;i<=num;i++)    if (in[i]==0) ans++;    printf("%d",ans);    return 0; }