Atcoder ARC061 E Snuke's Subway Trip

E - すぬけ君の地下鉄旅行 / Snuke's Subway Trip

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Time limit : 3sec / Memory limit : 256MB

Score : 600 points

Problem Statement

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≤105
• 0≤M≤2×105
• 1≤pi≤N (1≤i≤M)
• 1≤qi≤N (1≤i≤M)
• 1≤ci≤106 (1≤i≤M)
• pi≠qi (1≤i≤M)

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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.

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Sample Input 1

Copy

3 3
1 2 1
2 3 1
3 1 2

Sample Output 1

Copy

1

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

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Sample Input 2

Copy

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

Copy

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.

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Sample Input 3

Copy

2 0

Sample Output 3

Copy

-1

题目大意：已知某城市的地铁网由一些地铁线路构成，每一条地铁线路由某一个公司运营，该城市规定：若乘坐同一公司的地铁，从开始到换乘只需要一块钱，换乘其他公司的价格也是一块钱，问从1号地铁站到n号地铁站的最下价格。

思路：一眼看上去就是个图论题，我们考虑普通建图，无法保证是否换乘。考虑重新建图，我们对于每一个点，建一些“虚点”，对于每一个地铁网的联通块内部，路径为0，外部路径为1，再跑一个DFS即可。

#include
#include
#include
#include
#include
#include
#include
#include
#include
#define pa pair
 
using namespace std;

const int MAXN=1000010;
int dis[MAXN];
vector  g[MAXN];
vector  new_g[MAXN],all[MAXN];
int visit[MAXN],ptr[MAXN];
queue  q;
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++)
	{
		int u,v,c;
		scanf("%d%d%d",&u,&v,&c);
		u--,v--;
		g[u].push_back(make_pair(c,v));
		g[v].push_back(make_pair(c,u));
		all[c].push_back(u);
		all[c].push_back(v);
	}
	for (int i=0;i