codefroces 601A 题意唬人的最短路

题意：

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

input

output

-1

input

output

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.

N个点M条双向边，代表M条铁路，若两点之间无铁路相连，则建一条公路。从1到N点分别走两种路，若能保证汽车和火车在同一时刻不会在同一个点相遇（起始点除外），则输出两种路线都能到达N点的最小时刻。否则输出-1

分析：

两点之间只有一条边，所以无论怎么走都绝逼不可能中途相遇.......然后取两条最短路的较大值就可以了

代码：可以用一个矩阵临时存储一下输入的边

#include<bits/stdc++.h>
#include<queue>
using namespace std;
const int maxn=430;
const __int64 inf=(__int64)1<<62;//0x3f3f3f3f 太小了.... typedef pair<__int64,int> p;//<a,b> 起点到b的最短距离为a struct edge {int to;int cost;edge(int tto=0,int ccost=0):to(tto),cost(ccost){}
};int n,m;
int tmp[maxn][maxn];
__int64 d[maxn];
//int prev[maxn];
vector<edge> g[maxn];void dijkstra(int s) {priority_queue<p,vector<p>,greater<p> >qu;//堆按照p的first（最短距离）排序，小顶堆 fill(d,d+n+1,inf);
//  fill(prev,prev+n+1,-1);//+n 太小了...
//  memset(prev,sizeof(prev),-1);//混用会报错 d[s]=0;qu.push(p(0,s));//从起点出发到顶点s的最短距离为0while(!qu.empty()) {p temp=qu.top();qu.pop();int tmpv=temp.second;if(temp.first>d[tmpv]) continue;//跳过更新过程中入队的非最小值 for(int i=0;i<g[tmpv].size();++i) {//遍历该顶点连出的每条边 edge e=g[tmpv][i];if(d[e.to]>d[tmpv]+e.cost) {d[e.to]=d[tmpv]+(__int64)e.cost;
//              prev[e.to]=tmpv;qu.push(p(d[e.to],e.to));}}}
}//vector<int> getpath(int t) {//s -> t
//  vector<int> path;
//  while(t!=-1) {//从 t 倒着走，一直走到 s
//      path.push_back(t);
//      t=prev[t];
//  }
//  reverse(path.begin(),path.end());
//  return path;
//}int main() {//31 ms  8000 KBint a,b,c;scanf("%d%d",&n,&m);for(int i=0;i<=n;++i) {g[i].clear();}for(int i=1;i<=m;++i) {scanf("%d%d",&a,&b);tmp[a][b]=tmp[b][a]=1;}if(tmp[1][n]==1){//1和N铁路相连，走铁路最短路一定是1，只需要求公路的最短路 for(int i=1;i<=n;++i){for(int j=1;j<=n;++j){if(tmp[i][j]==0){g[i].push_back(edge(j,1));g[j].push_back(edge(i,1));                  }}} }else{for(int i=1;i<=n;++i){for(int j=1;j<=n;++j){if(tmp[i][j]==1){g[i].push_back(edge(j,1));g[j].push_back(edge(i,1));                 }}}         }dijkstra(1);if(d[n] == inf) puts("-1");else printf("%d\n",d[n]);return 0;
}

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