#include <bits/stdc++.h>
#define ll long long
#define R register
#define F(i,a,b) for(int i = (a);i<=(b);i++)
using namespace std;
inline int read(){R int x=0,t=1;R char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-') t=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*t;}
const int N=3e5+10;
int n,m,a[N],ans,top,dfn[N],low[N],l=1,r=0,st[N],cnt,b[N],f1[N],f2[N],u_[N],v_[N],S[N],q[N];
vector<int> g[N],G[N],G_[N],g1,g2;
bool vis[N];
void tarjan(int u)
{
	dfn[u]=low[u]=++top;
	st[++r]=u;
	vis[u]=1;
	for(int i = 0;i<g[u].size();i++){
		int v=g[u][i];
		if(!dfn[v]){
			tarjan(v);
			low[u]=min(low[u],low[v]);
		}
		else if(vis[v]){//返祖 
			low[u]=min(low[u],dfn[v]);
		}
	}
	if(dfn[u]==low[u]){
		cnt++;
		while(st[r]!=u){
			vis[st[r]]=0;
		//	cout << cnt << '\n';
			b[st[r]]=cnt;
			S[cnt]++;
			r--;
		}
		vis[u]=0;
		//cout << cnt << '\n';
		b[u]=cnt;
		S[cnt]++;
		r--;
	}
	return; 
}
struct Node
{
	int id,sum,wsr;
};
bool operator<(const Node&a,const Node&b)
{
	return a.sum<b.sum;
}

//priority_queue<Node>q;
/*
inline void dij(int rt)//dij跑缩后的最长路 
{
	dis[rt][0]=S[rt];
	q.push({rt,S[rt],0});
//	vis[rt]=1;
	for(;q.size();){
		Node u=q.top();
		q.pop();
	//	cout << u.id << " " << u.sum << " " << u.wsr << '\n';
		///if(vis[u.id]) continue;
		//vis[u.id]=1;
		for(int i = 0;i<G_[u.id].size();i++){
			int v=G_[u.id][i];
			if(u.wsr==0){
				if(dis[v][1]<dis[u.id][0]+S[v]){
					dis[v][1]=dis[u.id][0]+S[v];
					q.push({v,dis[v][1],1});
				}
			}
		}
		for(int i = 0;i<G[u.id].size();i++){
			int v=G[u.id][i];
			if(u.wsr!=0){
				if(dis[v][1]<dis[u.id][1]+S[v]){
					dis[v][1]=dis[u.id][1]+S[v];
					q.push({v,dis[v][1],1});
				}
			}
			else{
				if(dis[v][0]<dis[u.id][0]+S[v]){
					dis[v][0]=dis[u.id][0]+S[v];
					q.push({v,dis[v][0],0});
				}
				if(dis[v][1]<dis[u.id][1]+S[v]){
					dis[v][1]=dis[u.id][1]+S[v];
					q.push({v,dis[v][1],1});
				}
			}
		}
	}
}*/
inline void add(int ui,int vi)//原图 
{
	g[ui].push_back(vi);
	//g[vi].push_back(ui);
	return;
}
inline void add1(int ui,int vi)//缩完点的图 
{
	G[ui].push_back(vi);
	G_[vi].push_back(ui);
}
inline void solve()
{
	n=read(),m=read();
	F(i,1,m){
		u_[i]=read(),v_[i]=read();
		add(u_[i],v_[i]);
	}
	//dij(1);
	tarjan(1);
	F(i,1,n) if(b[i]==0) b[i]=++cnt,S[cnt]=1;
//	cout << l << " " << r << '\n'; 
//	F(i,1,n) cout << b[i] << '\n';
//	F(i,1,cnt)
//	{
//		cout << i << " " << S[i] << '\n';
//	}
	F(i,1,m){
		int ui=u_[i],vi=v_[i];
		if(b[ui]!=b[vi]){//两个不属于同一个强连通分量
			//bdcdcout << ui << " " << vi << '\n'; 
			//cout << b[ui] << " " << b[vi] << '\n'; 
			//cout << "------------------------\n";
			g1.push_back(ui);
			g2.push_back(vi);
			add1(ui,vi);
		}
	}
	l=1,r=0;
	q[++r]=b[1];
	f1[b[1]]=0;
	for(;l<=r;)
	{
		int u=q[l++];
		for(int i = 0;i<G[u].size();i++){
			int v=G[u][i];
			f1[v]=f1[u]+S[v];
			q[++r]=v;
		}
	}
	l=1,r=0;
	q[++r]=b[1];
	f2[b[1]]=1;
	for(;l<=r;)
	{
		int u=q[l++];
		for(int i = 0;i<G_[u].size();i++){
			int v=G_[u][i];
			f2[v]=f2[u]+S[v];
			q[++r]=v;
		}
	}
	int maxn=0;
	for(int i = 0;i<g1.size();i++){
		int ui=g1[i],vi=g2[i];
		maxn=max(maxn,f1[ui]+f2[vi]);
		maxn=max(maxn,f1[vi]+f2[ui]);
	}
	cout << maxn;
}
signed main()
{
	solve();
	return 0;
}

先缩点，再拓扑求正图和反图的到1的距离

今天模拟赛上搞了2h都只有7pts，赛后发现思路只对了一半，调完14pts，玄关求调