Snuke's Subway Trip: Dijkstra's Algorithm

Problem Statement （See: http://arc061.contest.atcoder.jp/tasks/arc061_c）

Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.

The i-th ( 1≤i≤M ) line connects station pi and qi bidirectionally. There is no intermediate station. This line is operated by company ci.

You can change trains at a station where multiple lines are available.

The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.

Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.

Constraints

• 2≤N≤10^5
• 0≤M≤2×10^5
• 1≤pi≤N (1≤i≤M)
• 1≤qi≤N (1≤i≤M)
• 1≤ci≤10^6 (1≤i≤M)
• pi≠qi (1≤i≤M)

---

Input

The input is given from Standard Input in the following format:

N M
p1 q1 c1
:
pM qM cM

Output

Print the minimum required fare. If it is impossible to get to station N by subway, print -1 instead.

Sample Input 1

3 3
1 2 1
2 3 1
3 1 2

Sample Output 1

1

Use company 1's lines: 1 → 2 → 3. The fare is 1 yen.

---

Sample Input 2

8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5

Sample Output 2

2

First, use company 1's lines: 1 → 3 → 2 → 5 → 6. Then, use company 5's lines: 6 → 7 → 8. The fare is 2 yen.

---

Sample Input 3

2 0

Sample Output 3

-1

#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll_t;
 
#define INF 1E9
 
ll_t pack(ll_t x, ll_t y) {
    return x * 1E9 + y;
}
 
using edge = pair<ll_t, int>;
 
struct Dist {
    ll_t v;
    int d;
    Dist(ll_t v_, int d_ = INF) : v(v_), d(d_) { } 
    bool operator < (const Dist& rhs) const {
        return d > rhs.d;
    }
};
 
int N;
int M;
unordered_map<ll_t, vector<edge>> graph;
unordered_map<ll_t, int> dist;
unordered_set<ll_t> done;
 
int main() {
    scanf("%d%d", &N, &M);
 
    for (int i = 1; i <= M; i++) {
        int p, q, c;
        scanf("%d%d%d", &p, &q, &c);
        graph[pack(p, c)].push_back(edge(pack(q, c), 0));
        graph[pack(q, c)].push_back(edge(pack(p, c), 0));
        
        graph[pack(p, c)].push_back(edge(pack(p, 0), 0));
        graph[pack(p, 0)].push_back(edge(pack(p, c), 1));
        graph[pack(q, c)].push_back(edge(pack(q, 0), 0));
        graph[pack(q, 0)].push_back(edge(pack(q, c), 1));
    }
 
    priority_queue<Dist> q;
    dist[pack(1, 0)] = 0;
    q.push(Dist(pack(1, 0), 0));
    while (!q.empty()) {
        Dist t = q.top();
        q.pop();
        if (done.count(t.v)) continue;
        done.insert(t.v);
 
        for (auto& kv : graph[t.v]) {
            ll_t u = kv.first;
            int d = kv.second;
            int tmp = dist[t.v] + d;
            if (dist.count(u) == 0 || tmp < dist[u]) {
                dist[u] = tmp;
                q.push(Dist(u, tmp));
            }
        }
    }
    if (dist.count(pack(N, 0)) == 0)
        printf("-1
");
    else
        printf("%d
", dist[pack(N, 0)]);
    return 0;
}