Problem Description
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 0001≤M≤200 0001≤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.
Example
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.
题意：n 个点 m 条边，只能移动两次，问能否从 1 号点到达 n 号点
思路：建图 dfs 即可
Source Program
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#define EPS 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
const int MOD = 1E9+7;
const int N = 200000+5;
const int dx[] = {-1,1,0,0,-1,-1,1,1};
const int dy[] = {0,0,-1,1,-1,1,-1,1};
using namespace std;
struct Edge{
int to;
int next;
}edge[N];
int n,m;
int tot,head[N];
void addEdge(int from,int to){
edge[++tot].to=to;
edge[tot].next=head[from];
head[from]=tot;
}
bool dfs(int s,int step){
if(s==n&&step==2)
return true;
for(int i=head[s];i!=-1;i=edge[i].next){
int to=edge[i].to;
if(dfs(to,step+1))
return true;
}
return false;
}
int main() {
memset(head,-1,sizeof(head));
scanf("%d%d",&n,&m);
for(int i=1;i<=m;i++){
int x,y;
scanf("%d%d",&x,&y);
addEdge(x,y);
}
bool flag=dfs(1,0);
if(flag)
printf("POSSIBLE\n");
else
printf("IMPOSSIBLE\n");
return 0;
}
