建议使用 stringstream 读取数据。

每条线路都说明线路中按顺序的任意两点可以一趟到达，依此建图即可。

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100'000;
const int MAXM = 200'000;
int n, m, s; // 点数、边数、出发点编号
struct Edge
{
    int v, w;
};
vector<Edge> e[MAXN + 5];
int dis[MAXN + 5];  // dis[i]: s->i 的最短路
bool vis[MAXN + 5]; // vis[i]: i 号点是否已经确定了

struct Point
{
    int id, dis;
};
bool operator<(const Point &x, const Point &y)
{
    return x.dis > y.dis;
}
priority_queue<Point> q;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    // 输入
    cin >> m >> n;
    stringstream ss;
    string s;
    getline(cin, s); // 吃一个n后面的换行
    for (int t = 1; t <= m; t++)
    {
        getline(cin, s);
        ss.clear();
        ss << s;
        int last = -1, now;
        vector<int> points;
        while (ss >> now)
            points.push_back(now);
        for (int i = 0; i < points.size(); i++)
        {
            for (int j = i + 1; j < points.size(); j++)
            {
                e[points[i]].push_back((Edge){points[j], 1});
                //cout << points[i] << " " << points[j] << " " << 1 << "\n";
            }
        }
    }
    // dijkstra
    memset(dis, 0x3f, sizeof(dis));
    memset(vis, false, sizeof(vis));
    dis[1] = 0;
    q.push((Point){1, 0});
    for (int t = 1; t <= n - 1; t++)
    {
        // 找到当前距离最近且未确定的点
        while (!q.empty() && vis[q.top().id])
            q.pop();
        if (q.empty())
            break;
        int u = q.top().id;
        // u可以确定距离为 dis[u] 了
        vis[u] = true;
        // 用 u 优化其他点
        for (int i = 0; i < e[u].size(); i++)
        {
            int v = e[u][i].v;
            int w = e[u][i].w;
            if (vis[v])
                continue;
            if (dis[u] + w < dis[v])
            {
                dis[v] = dis[u] + w;
                q.push((Point){v, dis[v]});
            }
        }
    }

    // 输出
    if (dis[n] == 0x3f3f3f3f)
        cout << "NO\n";
    else
        cout << dis[n] - 1 << "\n";
    return 0;
}