Age of Moyu

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1483 Accepted Submission(s): 451

Problem Description

Mr.Quin love fishes so much and Mr.Quin’s city has a nautical system,consisiting of N ports and M shipping lines. The ports are numbered 1 to N . Each line is occupied by a Weitian. Each Weitian has an identification number.

The i -th (1≤i≤M) line connects port Ai and Bi (Ai≠Bi) bidirectionally, and occupied by Ci Weitian (At most one line between two ports).

When Mr.Quin only uses lines that are occupied by the same Weitian, the cost is 1 XiangXiangJi. Whenever Mr.Quin changes to a line that is occupied by a different Weitian from the current line, Mr.Quin is charged an additional cost of 1 XiangXiangJi. In a case where Mr.Quin changed from some Weitian A 's line to another Weitian's line changes to Weitian A 's line again, the additional cost is incurred again.

Mr.Quin is now at port 1 and wants to travel to port N where live many fishes. Find the minimum required XiangXiangJi (If Mr.Quin can’t travel to port N , print −1 instead)

Input

There might be multiple test cases, no more than 20 . You need to read till the end of input.

For each test case,In the first line, two integers N (2≤N≤100000) and M (0≤M≤200000) , representing the number of ports and shipping lines in the city.

In the following m lines, each contain three integers, the first and second representing two ends Ai and Bi of a shipping line (1≤Ai,Bi≤N) and the third representing the identification number Ci (1≤Ci≤1000000) of Weitian who occupies this shipping line.

Output

For each test case output the minimum required cost. If Mr.Quin can’t travel to port N , output −1 instead.

Sample Input

3 3

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1 2 1

1 3 2

2 3 1

2 0

3 2

1 2 1

2 3 2

Sample Output

-1

题意：有n个点,m条边，如果存在边a-b-c,从a走到b，若ab的权值==bc的权值，则路径为1，反之路径为2。现在求从1到n的最短路；

思路：优先队列的迪杰特斯拉；

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
using namespace std;
vector<int>edge[100005];
vector<int>dis[100005];
int v[100005];
int d[100005];
struct Node
{
    int last;
    int w;
    int id;
}x[100005];
bool operator <(Node A,Node B)
{
    return (A.w>B.w);
}
void dij(int k)
{
    priority_queue<Node>q;
    memset(v,0,sizeof(v));
    q.push(x[k]);
    while(!q.empty())
    {
        Node temp=q.top();
        int index=temp.id;
        //cout<<index<<endl;
        if(v[index]==1)continue;
        q.pop();
        for(int i=0;i<edge[index].size();i++)
        {
            int dot=edge[index][i];
            int tempDis=d[index];
            if(x[index].last!=dis[index][i])
                tempDis=d[index]+1;
            if(d[dot]>tempDis)
            {
                d[dot]=tempDis;
                x[dot].last=dis[index][i];
                x[dot].w=d[dot];
                q.push(x[dot]);
            }
        }
    }
}
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1;i<=n;i++)d[i]=1000000+10,x[i].id=i;
        for(int i=1;i<=100000;i++)dis[i].clear(),edge[i].clear();
        for(int i=0;i<m;i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            edge[a].push_back(b);
            edge[b].push_back(a);
            dis[a].push_back(c);
            dis[b].push_back(c);
        }
        x[1].last=-1;
        d[1]=0;
        x[1].w=0;
        dij(1);
        if(d[n]==1000000+10)
            printf("-1\n");
        else
            printf("%d\n",d[n]);
    }
    return 0;
}

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