专栏首页饶文津的专栏【CodeForces 602C】H - Approximating a Constant Range(dijk)

【CodeForces 602C】H - Approximating a Constant Range(dijk)

Description

In Absurdistan, there are n towns (numbered 1 through n) and m bidirectional railways. There is also an absurdly simple road network — for each pair of different towns x and y, there is a bidirectional road between towns x and yif and only if there is no railway between them. Travelling to a different town using one railway or one road always takes exactly one hour.

A train and a bus leave town 1 at the same time. They both have the same destination, town n, and don't make any stops on the way (but they can wait in town n). The train can move only along railways and the bus can move only along roads.

You've been asked to plan out routes for the vehicles; each route can use any road/railway multiple times. One of the most important aspects to consider is safety — in order to avoid accidents at railway crossings, the train and the bus must not arrive at the same town (except town n) simultaneously.

Under these constraints, what is the minimum number of hours needed for both vehicles to reach town n (the maximum of arrival times of the bus and the train)? Note, that bus and train are not required to arrive to the town n at the same moment of time, but are allowed to do so.

Input

The first line of the input contains two integers n and m (2 ≤ n ≤ 400, 0 ≤ m ≤ n(n - 1) / 2) — the number of towns and the number of railways respectively.

Each of the next m lines contains two integers u and v, denoting a railway between towns u and v (1 ≤ u, v ≤ n, u ≠ v).

You may assume that there is at most one railway connecting any two towns.

Output

Output one integer — the smallest possible time of the later vehicle's arrival in town n. If it's impossible for at least one of the vehicles to reach town n, output  - 1.

Sample Input

Input

4 2 1 3 3 4

Output

2

Input

4 6 1 2 1 3 1 4 2 3 2 4 3 4

Output

-1

Input

5 5 4 2 3 5 4 5 5 1 1 2

Output

3

Hint

In the first sample, the train can take the route

and the bus can take the route

. Note that they can arrive at town4 at the same time.

In the second sample, Absurdistan is ruled by railwaymen. There are no roads, so there's no way for the bus to reach town 4.

因为任意一对城市之间都有一条直通的路,要么是铁路要么是公路,因此1到n城市一定有铁路或公路,是铁路,就再去找公路的最短路,否则就找铁路的最短路。

#include<stdio.h>

const int maxn=0x7fff;
long long n,m,s,e,t[405][405],u[405][405],dist[405];
void dijk(int v0,long long r[][405])
{
    bool b[405];
    for(int i=1; i<=n; i++)
    {
        dist[i]=r[v0][i];
        b[i]=false;
    }
    dist[v0] = 0;
    b[v0] = true;
    for(int i=2; i<=n; i++)
    {
        long long mindis=maxn;
        int u = v0;
        for(int j=1; j<=n; j++)
            if((!b[j]) && dist[j]<mindis)
            {
                u = j;
                mindis = dist[j];
            }
        b[u]=true;
        for(int j=1; j<=n; j++)
            if((!b[j]) && r[u][j]<maxn)
                if(dist[u] + r[u][j] < dist[j])
                    dist[j] = dist[u] + r[u][j];
    }
}
int main()
{
    scanf("%lld%lld",&n,&m);
    for(int i=0; i<=n; i++)
        for(int j=0; j<=n; j++)
        {
            t[i][j]=maxn;
            u[i][j]=1;
        }
    for(int i=0; i<m; i++)
    {
        scanf("%lld%lld",&s,&e);
        t[s][e]=t[e][s]=1;
        u[s][e]=u[e][s]=maxn;
    }
    if(u[1][n]==1)//road直达,铁路的最短路
        dijk(1,t);
    else
        dijk(1,u);
    if(dist[n]>=maxn)
        printf("-1\n");
    else
        printf("%lld\n",dist[n]);

    return 0;
}

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