把每个车站看作一个点，将这个车站相连的第一个边设为0，其余为1. 跑一边最短路，累加即可。

#include <bits/stdc++.h>
using namespace std;
int n, m, k, A, B, d[105][105];

int main() {
	memset (d, 0x3f, sizeof(d));
	cin >> n >> A >> B;
	for (int i = 0; i <= n; i++)
		d[i][i] = 0;
	for (int i = 1; i <= n; i++) {
		cin >> k;
		for (int j = 1; j <= k; j++) {
			cin >> m;
			if (j == 1)
				d[i][m] = 0;
			else
				d[i][m] = 1;
		}
	}
	for (int k = 1; k <= n; k++)
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				d[i][j] = min (d[i][j], d[i][k] + d[k][j]);
	if (d[A][B] == 0x3f3f3f3f)
		cout << -1;
	else
		cout << d[A][B];
	return 0;
}

把每个车站看作一个点，将这个车站相连的第一个边设为0，其余为1. 跑一边最短路，累加即可。

#include <bits/stdc++.h>
using namespace std;
int n, m, k, A, B, d[105][105];

int main() {
	memset (d, 0x3f, sizeof(d));
	cin >> n >> A >> B;
	for (int i = 0; i <= n; i++)
		d[i][i] = 0;
	for (int i = 1; i <= n; i++) {
		cin >> k;
		for (int j = 1; j <= k; j++) {
			cin >> m;
			if (j == 1)
				d[i][m] = 0;
			else
				d[i][m] = 1;
		}
	}
	for (int k = 1; k <= n; k++)
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				d[i][j] = min (d[i][j], d[i][k] + d[k][j]);
	if (d[A][B] == 0x3f3f3f3f)
		cout << -1;
	else
		cout << d[A][B];
	return 0;
}