Atcoder At Beginner Contest 068 C

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 a_i and Island b_i.

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≤a_i<b_i≤N
(a_i,b_i)≠(1,N)
If i≠j, (a_i,b_i)≠(a_j,b_j).

Input

Input is given from Standard Input in the following format:

N M
a₁ b₁
a₂ b₂
:
a_M b_M

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

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

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
");
        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
");
    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;
    }