https://vjudge.net/problem/Kattis-planestrainsbutnotautomobiles

题意：给一个有向图，火车可以由任意一个起点开始，每一个点只能经过一次，在坐火车的时候你可以选择坐飞机到另外一个点，求坐飞机的最小次数，以及求出可能在哪里坐飞机和降落。

样例：

 

思路：求坐飞机次数 就是裸的最小路径覆盖，然后bfs求出可能的起点和终点。

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <cstdlib>
#include <list>
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f3f3f3f
#define FILL(a,b) (memset(a,b,sizeof(a)))
#define lson rt<<1
#define rson rt<<1|1
#define lowbit(a) ((a)&-(a))
#define ios std::ios::sync_with_stdio(false);std::cin.tie(0);std::cout.tie(0);
#define fi first
#define sc second
#define pb push_back
#define endl '\n'
#define all(x) (x).begin(),(x).end()
using namespace std;
typedef long long  ll;
typedef unsigned long long  ull;
typedef pair<ll,ll> pii;
int dx[8]= {-1,1,0,0,1,1,-1,-1},dy[8]= {0,0,1,-1,-1,1,-1,1};
const ll mod=998244353;
const ll N =2e5+10;
const ll M =250000;
const double eps = 1e-4;
//const double pi=acos(-1);
ll re(){ll x;scanf("%lld",&x);return x;}
ll qk(ll a,ll b){ll ans=1;while(b){if(b&1) ans=ans*a%mod;a=a*a%mod;b/=2;}return ans;}
inline int read(){int sgn = 1; int sum = 0;char ch = getchar();while (ch < '0' || ch > '9') {if(ch == '-')   sgn = -sgn;ch = getchar();}while ('0' <= ch && ch <= '9') {sum = sum*10+(ch-'0');ch = getchar();}return sgn*sum;}ll n,m;
vector<int> g[N];
vector<int> a;
int match[N];
int vis[N];
int vis1[N];
bool dfs(int x){for(int i:g[x]){if(!vis1[i]){vis1[i]=1;if(!match[i]||dfs(match[i])){match[x]=i;match[i]=x;return 1;}}}return 0;
}
void bfs(){FILL(vis,0);queue<int> q;for(int i=1;i<=n;i++){if(!match[i]) q.push(i);}while(!q.empty()){int u=q.front();q.pop();if(vis[u]) continue;vis[u]=1;a.pb(u);for(int v:g[u]){if(v!=match[u]&&match[v]){q.push(match[v]);}}}for(int i=n+1;i<=2*n;i++){if(!match[i]) q.push(i);}while(!q.empty()){int u=q.front();q.pop();if(vis[u]) continue;vis[u]=1;a.pb(u-n);//cout<<u-n<<endl;for(int v:g[u]){if(v!=match[u]&&match[v]){q.push(match[v]);}}}sort(all(a));a.erase(unique(all(a)),a.end());
}
void sovle(){n=re(),m=re();for(int i=1;i<=m;i++){int u=re(),v=re();g[u].pb(v+n);g[v+n].pb(u);}ll ans=0;for(int i=1;i<=n;i++){FILL(vis1,0);if(dfs(i)) ans++;}bfs();cout<<n-ans-1<<endl;if(n-ans-1!=0)for(int i:a) cout<<i<<" ";
}int main()
{iosint t=1;while(t--){sovle();}return 0;
}