其实最阴险的就是：合法的赛道的条件中只说了终点不能往回走，没说起点不能从其他点可达！
估计很多人想当然的对称思维就会认为后面的条件也自然地满足了（我就是QAQ）
第一个subtask没有任何影响，但是第二个split的时猴会把带自环的点排除掉
#include <iostream>
#include <stdio.h>
#include <queue>
using namespace std;

bool adj[55][55] = {0};
int adjList[55][55];
int adjNum[55] = {0};

int main() {
	int cnt = 0;
	while(1){
		int temp;
		scanf("%d", &temp);
		if(temp == -1) break;
		adj[cnt][temp] = 1;
		adjList[cnt][adjNum[cnt]] = temp;
		adjNum[cnt]++;
		while(1){
			int tmp;
			scanf("%d", &tmp);
			if(tmp == -2) break;
			adj[cnt][tmp] = 1;
			adjList[cnt][adjNum[cnt]] = tmp;
			adjNum[cnt]++;
		}
		cnt++;
	}
	int N = cnt;
	//cout << N << endl;
	bool conn[2][55][55];
	for(int i = 0; i <= N; i++){
		for(int j = 0; j <= N; j++){
			if(i == j) conn[0][i][j] = 1;
			else conn[0][i][j] = adj[i][j];
		}
	}
	//floyd厕连通性
	for(int i = 1; i < N; i++){
		int from = (i+1)%2, to = i%2;
		for(int j = 0; j <= N; j++){
			for(int k = 0; k <= N; k++){
				if(conn[from][j][k]){
					conn[to][j][k] = 1;
					continue;
				}
				if(conn[from][j][i] && conn[from][i][k]){
					conn[to][j][k] = 1;
				}
				else conn[to][j][k] = 0;
			}
		}
	}
	//bfs看是否必需
	int uavNum = 0;
	int uav[55];
	int split[55];
	int splitNum = 0;
	int tar = (N+1)%2;
	for(int i = 1; i < N; i++){
		bool used[55] = {1,0};
		queue<int> bfs;
		bfs.push(0);
		used[i] = 1;
		bool kydd = 0;
		while(!bfs.empty()){
			int t = bfs.front();
			bfs.pop();
			for(int j = 0; j < adjNum[t]; j++){
				if(used[adjList[t][j]]) continue;
				used[adjList[t][j]] = 1;
				if(adjList[t][j] == N){
					kydd = 1;
					goto done;
				}
				bfs.push(adjList[t][j]);
			}
		}
		done:
		if(!kydd){
			uav[uavNum] = i;
			uavNum ++;
			used[i] = 0;
			bool spl = 1;
			for(int j = 0; j <= N; j++){
				for(int k = 0; k <= N; k++){
					if(used[j] && !used[k] && conn[tar][k][j]){
						spl = 0;
						goto done1;
					}
				}
			}
			done1:
			if(spl){
				split[splitNum] = i;
				splitNum ++;
			}
		}
	}
	printf("%d", uavNum);
	for(int i = 0; i < uavNum; i++){
		printf(" %d", uav[i]);
	}
	printf("\n");


	printf("%d", splitNum);
	for(int i = 0; i < splitNum; i++){
		printf(" %d", split[i]);
	}
	printf("\n");
	return 0;
}