Snuke's Subway Trip

すぬけ君の地下鉄旅行 / Snuke's Subway Trip

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)

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

Sample Output 1

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

分析：根据贪心思想，从一个点到另一个点必然是从之前点最小花费转移过来的，否则不会更优；

　　　所以每个点维护最小花费和到达前的公司id，用最短路拓展即可；

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=1e5+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int n,m,k,t,h[maxn],tot,ans[maxn];
set<int>comp[maxn];
struct node
{
    int to,nxt,com;
}e[maxn<<2];
void add(int x,int y,int z)
{
    tot++;
    e[tot].to=y;
    e[tot].com=z;
    e[tot].nxt=h[x];
    h[x]=tot;
}
priority_queue<pair<int,int> >p;
int main()
{
    int i,j;
    memset(ans,inf,sizeof ans);
    scanf("%d%d",&n,&m);
    while(m--)
    {
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        add(a,b,c);
        add(b,a,c);
    }
    p.push({0,1});ans[1]=0;
    while(!p.empty())
    {
        int now=p.top().se,ca=-p.top().fi;
        p.pop();
        if(ans[now]<ca)continue;
        for(i=h[now];i;i=e[i].nxt)
        {
            int to=e[i].to,to_com=e[i].com;
            int to_ca=ca+(!comp[now].count(to_com));
            if(ans[to]>to_ca)
            {
                ans[to]=to_ca;
                p.push({-to_ca,to});
                comp[to].clear();
                comp[to].insert(to_com);
            }
            else if(ans[to]==to_ca)
            {
                comp[to].insert(to_com);
            }
        }
    }
    printf("%d
",ans[n]==inf?-1:ans[n]);
    //system("Pause");
    return 0;
}