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"; } }