Cat Snuke and a Voyage --AtCoder

题目描述

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).

输入

Input is given from Standard Input in the following format:
N M
a1 b1
a2 b2
:
aM bM

输出

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

样例输入

3 2
1 2
2 3

样例输出

POSSIBLE

分析：反正只需要两次即可，直接判断第一个节点后和第n个节点前有没有通。

#include <iostream>
#include <string>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#define range(i,a,b) for(int i=a;i<=b;++i)
#define LL long long
#define rerange(i,a,b) for(int i=a;i>=b;--i)
#define fill(arr,tmp) memset(arr,tmp,sizeof(arr))
using namespace std;
int n,m,a,b;
bool vis[],flag;
void init(){
    cin>>n>>m;
    fill(vis,);
    range(i,,m){
        cin>>a>>b;
        if(flag)continue;
        if(a==){
            flag=vis[b];
            vis[b]=true;
        }
        else if(b==n){
            flag=vis[a];
            vis[a]=true;
        }
    }
}
void solve(){
    cout<<(flag?"POSSIBLE":"IMPOSSIBLE")<<endl;
}
int main() {
    init();
    solve();
    return ;
}