Problem Description

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)

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.

Example

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

題意：n 個點 m 條邊，每條邊給出鄰接點的同時給出一個價值，代表的是該邊所屬的公司，現在要從 1 號點到 n 號點，已知在同一個公司上行走時，無需付出任何代價，但當走到令一公司的邊上時，需要花費 1 元，求最小花費

思路：

這個題的問題在于讓相同公司的道路之間行走花費為 0，不同公司之間的道路行走花費為 1，于是整張圖的每個公司可以看做一個虛點，其連接著其所屬的邊的兩點

由于在同一個公司的路上行走時，不耗費任何價值，那么可以對同一公司的路上進行拆點建圖，即對于所屬公司為 val 的邊 x-y，將其拆成 x-newX、newX-newY、newY-y 三條邊，其價值分別為 1、0、1，然后對整個圖跑 SPFA 求最短路即可

跑完 SPFA 時，由于在起點時需要一個初始代價，中間行走時不需要代價，離開時也需要一個代價，因此 dis[n]/2 就是答案

此外，重點在于如何拆點能保證點不重復，考慮到 map 的特性，我們令 mp[a][b] 的值從 n 號點開始遞增，從而保證點的唯一性

需要注意的是，由于拆完點后點和邊的數目會很多，因此數組范圍要開的大一點

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #define EPS 1e-9 #define PI acos(-1.0) #define INF 0x3f3f3f3f #define LL long long const int MOD = 1E9+7; const int N = 1000000+5; const int dx[] = {0,0,-1,1,-1,-1,1,1}; const int dy[] = {-1,1,0,0,-1,1,-1,1}; using namespace std; struct Edge {int to,next,val;Edge(){}Edge(int to,int next,int val):to(to),next(next),val(val){}bool operator < (const Edge& rhs)const{return val>rhs.val;} }edge[N*2]; int tot,head[N]; void addEdge(int from,int to,int val) {edge[tot].next=head[from];edge[tot].to=to;edge[tot].val=val;head[from]=tot++; } bool vis[N]; int dis[N]; int SPFA(int s,int e) {memset(vis,false,sizeof(vis));memset(dis,INF,sizeof(dis));dis[s]=0;priority_queue<Edge> Q;Q.push(Edge(s,0,0));while (!Q.empty()) {Edge temp=Q.top();Q.pop();if(vis[temp.to])continue;vis[temp.to]=true;for(int i=head[temp.to]; i!=-1; i=edge[i].next) {if(dis[edge[i].to]>dis[temp.to]+edge[i].val) {dis[edge[i].to]=dis[temp.to]+edge[i].val;Q.push(Edge(edge[i].to,0,dis[edge[i].to]));}}}if(dis[e]!=INF)return dis[e]/2;return -1; } map<int,int> mp[N]; int allPoint; int getPoint(int x,int y) {if (!mp[x][y])mp[x][y]=++allPoint;return mp[x][y]; } int main() {int n,m;scanf("%d%d",&n,&m);memset(head,-1,sizeof(head));allPoint=n;for (int i=1; i<=m; i++) {int x,y,w;scanf("%d%d%d",&x,&y,&w);int newX=getPoint(x,w);int newY=getPoint(y,w);addEdge(x,newX,1);addEdge(newX,x,1);addEdge(newX,newY,0);addEdge(newY,newX,0);addEdge(y,newY,1);addEdge(newY,y,1);}int res=SPFA(1,n);printf("%d\n",res);return 0; }

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