codeforces 601A The Two Routes（BFS）

题目链接

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.

Sample test(s)

Input

4 2
1 3
3 4

Output

2

Input

4 6
1 21 31 4
2 32 4
3 4

Output

-1

Input

5 5
4 23 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.

题意：n个城镇，两个城镇之间要么有一条铁路，要么有一条公路，通过一条铁路和一条公路的时间均为1。有一辆汽车和一辆火车，他们同时从点1出发，要到达点n，中间不会停留。为了避免交通事故，他们不能在同一时刻到达同一个点（除开终点n）。问两辆车都到达n点的最短时间。

题解：如果1到n是铁路相连，那么火车一定走这条路，汽车一定是走最短路。反之同理。

代码如下：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<set>
typedef __int64 LL;
typedef unsigned __int64 LLU;
const int nn=410;
const int inf=0x3fffffff;
const LL inf64=(LL)inf*inf;
const double eps = 1e-12;
using namespace std;
int n,m;
struct node
{
int k;
int en,next;
}E[nn*nn*2];
int p[nn],num;
void init()
{
memset(p,-1,sizeof(p));
num=0;
}
void add(int st,int en,int k)
{
E[num].k=k;
E[num].en=en;
E[num].next=p[st];
p[st]=num++;
}
vector<int>ve[nn];
bool tu[nn][nn];
int dis[nn];
int dp[nn];
queue<int>que;
int bfs(int k)
{
int i;
memset(dis,-1,sizeof(dis));
dis[1]=0;
que.push(1);
while(que.size())
{
int sta=que.front();
que.pop();
for(i=p[sta];i+1;i=E[i].next)
{
if(E[i].k==k)
{
int w=E[i].en;
if(dis[w]==-1)
{
dis[w]=dis[sta]+1;
que.push(w);
}
}

}
}
if(dis
==-1)
return inf;

return dis
;
}
int solve()
{
int re=max(bfs(0),bfs(1));
if(re==inf)
return -1;
return re;
}
int main()
{
int i,j;
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(tu,false,sizeof(tu));
init();
for(i=1;i<=m;i++)
{
int u,v;
scanf("%d%d",&u,&v);
tu[u][v]=tu[v][u]=true;
}
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
if(i==j)
continue;
if(tu[i][j])
{
add(i,j,0);
}
else
add(i,j,1);
}
}
printf("%d\n",solve());
}
return 0;
}