最短路径

最近写了挺多的最短路径的题目，我觉得有必要总结一下了。

首先一道入门的水题：

A. The Two Routes

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 y if 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.

Examples

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

Note

In the first sample, the train can take the route  and the bus can take the route . Note that they can arrive at town 4 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.

这道题目的意思就是铁路与公路不能共存，问火车与汽车花费的最长时间，如果有一种交通方式不能到达目的地，则输出-1，这题我用了dijkstra算法做

#include<iostream>
#include<cmath>
#include<cstring>
#define inf 0x3f3f3f3f
using namespace std;
int m,n;
int vis[450],vis1[450];
int a[450][450],b[450][450],d[450],d1[450];
void dijkstra(){
    int i,j,minn,flag;
    memset(vis,0,sizeof(vis));
    vis[1]=1;
    for(i=2;i<=m;i++){
        d[i]=a[1][i];
    }
    for(i=1;i<=m;i++){
        minn=inf;
        for(j=1;j<=m;j++){
            if(!vis[j]&&d[j]<minn){
                minn=d[j];
                flag=j;
    //            cout<<d[j]<<" "<<minn<<endl;
            }
        }
    //    cout<<minn<<endl;
        if(minn==inf){
            break;
        }
    //    cout<<minn<<endl;
        vis[flag]=1;
    //    cout<<d[flag]<<endl;
    //cout<<a[flag][4];
    //cout<<vis[4];
        for(j=1;j<=m;j++){
            if(!vis[j]&&(d[j]>d[flag]+a[flag][j])){
            //    cout<<d[j]<<" "<<d[flag]+a[flag][j]<<endl;
            //    cout<<flag<<" "<<j<<endl;
            //    cout<<d[3]<<" "<<a[3][2]<<endl;
                d[j]=d[flag]+a[flag][j];
            //    cout<<d[j]<<"*"<<j<<endl;
            }
        }
    //    cout<<d[flag]<<"*"<<endl;
    }
}
void dijkstra1(){
    int i,j,minn,flag;
    memset(vis1,0,sizeof(vis1));
    vis1[1]=1;
    for(i=2;i<=m;i++){
        d1[i]=b[1][i];
    }
    for(i=1;i<=m;i++){
        minn=inf;
        for(j=1;j<=m;j++){
            if(!vis1[j]&&d1[j]<minn){
                minn=d1[j];
                flag=j;
            }
        }
        if(minn==inf){
            break;
        }
        vis1[flag]=1;
        for(j=1;j<=m;j++){
            if(!vis1[j]&&d1[j]>(d1[flag]+b[flag][j])){
                d1[j]=d1[flag]+b[flag][j];
            }
        }
    }
}
int main()
{
    int i,j,x,y;
    while(cin>>m>>n){
        memset(a,inf,sizeof(a));
        memset(d,inf,sizeof(d));
        memset(d1,inf,sizeof(d1));
        for(i=0;i<n;i++){
            cin>>x>>y;
            a[x][y]=1;
            a[y][x]=1;
        }
        for(i=1;i<=m;i++){
            for(j=1;j<=m;j++){
                if(a[i][j]==1){
                    b[i][j]=inf;
                }
                else{
                    b[i][j]=1;
                }
    //            cout<<b[i][j]<<" ";
            }
        }
        dijkstra();
        dijkstra1();
    //    cout<<d[m]<<endl<<d1[m]<<endl;
        if(d[m]==inf||d1[m]==inf){
            cout<<"-1"<<endl;
        }
        else{
            cout<<max(d[m],d1[m])<<endl;
        }
    }
    return 0;
}

改bug 用了很长时间，就是把memset初始化错了memset对于整数只能初始化为0，而不能初始化为其他的数，我就是因为初始化为1 找了很久的错误，找bug的痕迹也留下来了，警醒自己吧，哈哈。