Atcoder At Beginner Contest 068 C - Cat Snuke and a Voyage

C - Cat Snuke and a Voyage

Time limit : 2sec / Memory limit : 256MB

Score : 300 points

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

Copy

3 2
1 2
2 3

Sample Output 1

Copy

POSSIBLE

Sample Input 2

Copy

4 3
1 2
2 3
3 4

Sample Output 2

Copy

IMPOSSIBLE

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

Sample Input 3

Copy

100000 1
1 99999

Sample Output 3

Copy

IMPOSSIBLE

Sample Input 4

Copy

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

Sample Output 4

Copy

POSSIBLE

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

bfs搜索

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
vector<int>v[200006];
int n,m,x,y,vis[200006];
bool flag;
void bfs(int x,int y,int k)
{
if(x==y && k<=2)
{
printf("POSSIBLE\n");
flag=true;
return;
}
if(k>2) return;
vis[x]=1;
for(int i=0;i<v[x].size();i++)
{
if(!vis[v[x][i]])
{
vis[v[x][i]]=1;
bfs(v[x][i],y,k+1);
vis[v[x][i]]=0;
}
}
vis[x]=0;
}
int main()
{
scanf("%d%d",&n,&m);
flag=false;
memset(vis,0,sizeof(vis));
while(m--)
{
scanf("%d%d",&x,&y);
v[x].push_back(y);
v[y].push_back(x);
}
bfs(1,n,0);
if(!flag) printf("IMPOSSIBLE\n");
return 0;
}

再来个set简单方法，因为此题最多找两步，如果说步数更多的话就只能搜索了

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
set<int>s,v;
int n,m,x,y;
int main()
{
scanf("%d%d",&n,&m);
s.insert(1);
v.insert(n);
while(m--)
{
scanf("%d%d",&x,&y);
if(x==1) s.insert(y);
if(y==1) s.insert(x);
if(x==n) v.insert(y);
if(y==n) v.insert(x);
}
for(set<int>::iterator it=s.begin();it!=s.end();it++)
{
if(v.count(*it)==1)
{
puts("POSSIBLE");
return 0;
}
}
puts("IMPOSSIBLE");
return 0;
}