题意
有一个有向图，允许最多走一次逆向的路，问从1再走回1，最多能经过几个点。
思路
（一）
首先先缩点。自己在缩点再建图中犯了错误，少连接了大点到其他点的边。
跑两次最长路，一次以1为起点，一次以1为终点（跑一遍反图）
然后枚举边，判断可否形成一个环。
（二）
分层图的思想
以为只有一次逆向的机会，可以建两层图，第一层向第二层连翻转的情况。
#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert> /* ⊂_ヽ
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( (ヽ
| |、＼
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'ノ ) Lﾉ */ using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3; //priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n' #define boost ios::sync_with_stdio(false);cin.tie(0)
#define rep(a, b, c) for(int a = (b); a <= (c); ++ a)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c); const ll oo = 1ll<<;
const ll mos = 0x7FFFFFFF; //
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //
const int mod = 1e9+;
const double esp = 1e-;
const double PI=acos(-1.0);
const double PHI=0.61803399; //黄金分割点
const double tPHI=0.38196601; template<typename T>
inline T read(T&x){
x=;int f=;char ch=getchar();
while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
return x=f?-x:x;
} inline void cmax(int &x,int y){if(x<y)x=y;}
inline void cmax(ll &x,ll y){if(x<y)x=y;}
inline void cmin(int &x,int y){if(x>y)x=y;}
inline void cmin(ll &x,ll y){if(x>y)x=y;} /*-----------------------showtime----------------------*/
const int maxn = 1e5+;
vector<int>mp1[maxn],mp2[][maxn];
int dfn[maxn],low[maxn],vis[maxn],gtot,nn,dp[maxn];
int col[maxn],vv[maxn];
stack<int>st; void tarjan(int u){
dfn[u] = low[u] = ++gtot;
st.push(u); vis[u] = ;
for(int i=; i<mp1[u].size(); i++){
int v = mp1[u][i];
if(dfn[v] == ) {
tarjan(v);
low[u] = min(low[u], low[v]);
}
else if(vis[v]){
low[u] = min(low[u], dfn[v]);
}
}
if(low[u] == dfn[u]){
int x;nn++;
while(!st.empty()){
int x = st.top(); st.pop();
col[x] = nn;
vis[x] = ;
dp[nn]++;
if(x == u) break;
}
}
}
int ans = ;
int dis[maxn][];
pii edge[maxn];
void dji(int s,int id){
priority_queue<pii>que; dis[s][id] = dp[s];
que.push(pii(dis[s][id], s));
while(!que.empty()){
int u = que.top().se; que.pop(); for(int i=; i<mp2[id][u].size(); i++) {
int v = mp2[id][u][i],w = dp[v]; if(dis[v][id] < dis[u][id] + w){
dis[v][id] = dis[u][id] + w;
que.push(pii(dis[v][id], v));
}
}
}
}
int main(){
int n,m;
scanf("%d%d", &n, &m);
rep(i, , m) {
int x,y; scanf("%d%d", &x, &y);
mp1[x].pb(y);
edge[i] = pii(x, y);
}
for(int i=; i<=n; i++)
if(!dfn[i])tarjan(i); int s = col[];
int pp = ;
memset(vis, , sizeof(vis));
for(int i=; i<=n; i++){
int u = col[i];
if(u == ) continue; for(int j=; j<mp1[i].size(); j++){
int v = col[mp1[i][j]];
if(v == || u == v) continue; mp2[][u].pb(v);
mp2[][v].pb(u);
}
}
dji(s, ); dji(s, ); int ans = dp[s];
for(int i=; i<=m; i++){
int u = col[edge[i].fi],v = col[edge[i].se];
if(dis[v][] && dis[u][])ans = max(ans, dis[v][] + dis[u][] - dp[s]);
} printf("%d\n", ans);
return ;
}