成功题解：https://ycwy-zd-rz.blog.luogu.org/cogs152-post
其实就是一道bfs水题，加了freopen后基本上都会过.
[cpp]#include<bits/stdc++.h>
using namespace std;
#define s freopen("mud.in","r",stdin)
#define ss freopen("mud.out","w",stdout)
#define sss fclose(stdin)
#define ssss fclose(stdout)
#define mint int// long
#define pd if(nx<=maxx && ny<=maxx && nx>=0 && ny>=0 && !mp[nx][ny])
mint mp[1010][1010];
mint rock,ax,ay;
mint maxx=1010;
mint hd,tl;
mint dx[]={0,0,0,1,-1};
mi[][/cpp]

https://ycwy-zd-rz.blog.luogu.org/cogs3008-post
//爱八卦的小朋友们
//爱八卦的小朋友们
[code=cpp]#include<bits/stdc++.h>
using namespace std;
#define mint long long int
mint n,m,ans;
bool f[10001];
struct u{
mint x,y;
};
u n[10001];
int main(){
freopen("friendscircle.in","r",stdin);
freopen("friendscircle.out","w",stdout);
cin>>n>>m;
int a=1;
f[a]=true;
for(mint i=1;i<=m;i++){
cin>>n[i].x>>n[i].y;
}
for(mint j=1;j<=n;j++)
for(mint i=1;i<=m;i++){
if(f[n[i].x] ) f[n[i].y]=true;
else if(f[n[i].y] ) f[n[