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 ai and Island bi.
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≤ai<bi≤N
 - (ai,bi)≠(1,N)
 - If i≠j, (ai,bi)≠(aj,bj).
 
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";
	}
 } 
                    
                
				 
                
            
        
浙公网安备 33010602011771号