Cat Snuke and a Voyage AtCoder

Problem Statement

 

In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N.

There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i.

Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services.

Help him.

Constraints

 

• 3≤N≤200 000
• 1≤M≤200 000
• 1≤a_i<b_i≤N
• (a_i,b_i)≠(1,N)
• If i≠j, (a_i,b_i)≠(a_j,b_j).

Input

 

Input is given from Standard Input in the following format:

N M
a1 b1
a2 b2
:
aM bM

Output

 

If it is possible to go to Island N by using two boat services, print POSSIBLE; otherwise, print IMPOSSIBLE.

Sample Input 1

 

3 2
1 2
2 3

Sample Output 1

 

POSSIBLE

Sample Input 2

 

4 3
1 2
2 3
3 4

Sample Output 2

 

IMPOSSIBLE

You have to use three boat services to get to Island 4.

Sample Input 3

 

100000 1
1 99999

Sample Output 3

 

IMPOSSIBLE

Sample Input 4

 

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

Sample Output 4

 

POSSIBLE

You can get to Island 5 by using two boat services: Island 1 -> Island 4 -> Island 5.

关于划船问题，从1道n只能转一次船也必须转一次船，所以需要有【1，c】，【c，n】两条船，所以只需要保存a=1的值和b=n的值。

上代码：

#include<iostream>
#include<cmath>
#include <algorithm>
using namespace std;
int main()
{
	int arr[200000];
	int brr[200000];
	int n,m;
	cin>>n>>m;
	int p=0,q=0;
	for(int i=0;i<m;i++)
	{
		int a,b;
		cin>>a>>b;
		if(a==1)
		{
			arr[p++]=b;
		}
		if(b==n)
		{
			brr[q++]=a;
		}
	}
	if(p==0||q==0)
	{
		cout<<"IMPOSSIBLE";
	}
	else
	{
		sort(arr,arr+p);
		sort(brr,brr+q);
		int i,j=1;
		for(i=0;i<p;i++)
		{
			j--;
			for(j;j<q;j++)
			{
				if(arr[i]==brr[j])
				{
					break;
				}
				if(arr[i]<brr[j])
					break;
			}
			if(arr[i]==brr[j])
			break;
		}
		if(arr[i]==brr[j]&&i!=p)
			cout<<"POSSIBLE";
		else
			cout<<"IMPOSSIBLE";
	}
 }