time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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.

【题目链接】:http://codeforces.com/contest/602/problem/C

【题解】



铁路和公路中必然有一种有一条从1->n的边

则那个人先走到n等着.

然后另外一个人跑最短路就可以了;

答案就是跑最短路的那个人用的时间.



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x) typedef pair<int,int> pii;
typedef pair<LL,LL> pll; const int MAXN = 4e2+100;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0); int n,m;
vector <int> G[2][MAXN];
bool flag[MAXN][MAXN];
queue <int> dl;
int dis[MAXN]; int main()
{
//freopen("F:\\rush.txt","r",stdin);
rei(n);rei(m);
rep1(i,1,m)
{
int x,y;
rei(x);rei(y);
flag[x][y] = flag[y][x] = true;
G[0][x].pb(y);
G[0][y].pb(x);
}
rep1(i,1,n)
rep1(j,1,n)
if (i!=j && !flag[i][j])
G[1][i].pb(j);
int bo = 0;
if (flag[1][n])
bo = 1;
memset(dis,255,sizeof dis);
dl.push(1);
dis[1] = 0;
while (!dl.empty())
{
int x = dl.front();
dl.pop();
for (auto y:G[bo][x])
{
if (dis[y]==-1 || dis[y] > dis[x]+1)
{
dis[y] = dis[x] + 1;
dl.push(y);
}
}
}
cout << dis[n]<<endl;
return 0;
}

【50.00%】【codeforces 602C】The Two Routes的更多相关文章

  1. 【50.00%】【codeforces 747C】Servers

    time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...

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

    Description through n) and m bidirectional railways. There is also an absurdly simple road network — ...

  3. 【 BowWow and the Timetable CodeForces - 1204A 】【思维】

    题目链接 可以发现 十进制4 对应 二进制100 十进制16 对应 二进制10000 十进制64 对应 二进制1000000 可以发现每多两个零,4的次幂就增加1. 用string读入题目给定的二进制 ...

  4. 【25.00%】【codeforces 584E】Anton and Ira

    time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...

  5. 【50.88%】【Codeforces round 382B】Urbanization

    time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...

  6. 【74.00%】【codeforces 747A】Display Size

    time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...

  7. 【codeforces 750A】New Year and Hurry

    time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...

  8. [CodeForces - 1225E]Rock Is Push 【dp】【前缀和】

    [CodeForces - 1225E]Rock Is Push [dp][前缀和] 标签:题解 codeforces题解 dp 前缀和 题目描述 Time limit 2000 ms Memory ...

  9. 【codeforces 709D】Recover the String

    [题目链接]:http://codeforces.com/problemset/problem/709/D [题意] 给你一个序列; 给出01子列和10子列和00子列以及11子列的个数; 然后让你输出 ...

随机推荐

  1. [React] Use the URL as the source of truth in React

    In Single Page Apps we're used to fetch the data on event callbacks. That disables the capacity to u ...

  2. Cocos2d-x 之大牛看法

    (未完毕) cocos2d-x并非一个适合网游client(mmo)的游戏引擎.越是大型游戏,这个小引擎就越无法驾驭(尽管它很受欢迎). 之前我在原来的公司使用的是自主研发的C3引擎,已经对外开放(尚 ...

  3. inflater-布局转化实现

    通俗的说,inflate就相当于将一个xml中定义的布局找出来.因为在一个Activity里如果直接用findViewById()的话,对应的是setConentView()的那个layout里的组件 ...

  4. 使用Notepad++的XML Tools插件格式化XML文件

    转自“”:https://blog.csdn.net/qq_36279445/article/details/79803310 1. 安装XML Tools插件 (1) 通过网址http://sour ...

  5. 洛谷P1143 进制转换

    题目描述 请你编一程序实现两种不同进制之间的数据转换. 输入输出格式 输入格式: 输入数据共有三行,第一行是一个正整数,表示需要转换的数的进制n(2≤n≤16),第二行是一个n进制数,若n>10 ...

  6. jQuery返回值:jQuery对象

    $(function(){ //返回值 alert($); //jQuery //以下返回的全是jQuery对象 alert($()); alert($('#box')); alert($('#box ...

  7. Codefroces Round#427 div2

    A. Key races time limit per test 1 second memory limit per test 256 megabytes input standard input o ...

  8. 关于Promise详解

    异步回调 回调地狱 在需要多个操作的时候,会导致多个回调函数嵌套,导致代码不够直观,就是常说的回调地狱 并行结果 如果几个异步操作之间并没有前后顺序之分,但需要等多个异步操作都完成后才能执行后续的任务 ...

  9. Unity5中的粒子缩放(附测试源码)

    本文章由cartzhang编写,转载请注明出处. 所有权利保留. 文章链接:http://blog.csdn.net/cartzhang/article/details/49363241 作者:car ...

  10. 【例题 7-14 UVA-1602】Lattice Animals

    [链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 借鉴网上的题解的. 思路是. 用"标准化"的思想. 确定基准点(0,0) 然后假设(0,0)是第一个连通块. 然 ...