思路：
n^2枚举（必须要n^2枚举啊）+拆点

特此嘲讽网上诸多垃圾题解，你们许多都是错的 —yyh

//By SiriusRen
#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define N 5555
int n,m,xx,yy,inf=0x3fffff,ans,ed=105;
struct Node{int x,y;}point[N];
struct Dinic{int first[107],next[N],v[N],w[N],tot,vis[107];void solve(int x,int y){memset(first,-1,sizeof(first)),tot=0;add(0,x,inf),add(x,x+n,inf),add(y,y+n,inf),add(y+n,105,inf);for(int i=1;i<=n;i++)add(i,i+n,1);for(int i=1;i<=m;i++)add(point[i].x+n,point[i].y,inf),add(point[i].y+n,point[i].x,inf);ans=min(ans,x=flow());}void add(int x,int y,int z){Add(x,y,z),Add(y,x,0);}void Add(int x,int y,int z){w[tot]=z,v[tot]=y,next[tot]=first[x],first[x]=tot++;}bool tell(){memset(vis,-1,sizeof(vis)),vis[0]=0;queue<int>q;q.push(0);while(!q.empty()){int t=q.front();q.pop();for(int i=first[t];~i;i=next[i])if(vis[v[i]]==-1&&w[i])vis[v[i]]=vis[t]+1,q.push(v[i]);}return vis[ed]!=-1;}int zeng(int x,int y){if(x==ed)return y;int r=0;for(int i=first[x];~i&&y>r;i=next[i])if(vis[v[i]]==vis[x]+1&&w[i]){int t=zeng(v[i],min(y-r,w[i]));w[i]-=t,w[i^1]+=t,r+=t;}if(!r)vis[x]=-1;return r;}int flow(){int jy=0,tmp;while(tell())while(tmp=zeng(0,inf))jy+=tmp;return jy;}
}dinic;
int main(){while(~scanf("%d%d",&n,&m)){ans=inf;for(int i=1;i<=m;i++){scanf(" (%d,%d)",&point[i].x,&point[i].y);point[i].x++,point[i].y++;}for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)if(i!=j)dinic.solve(i,j);if(ans==inf)printf("%d\n",n);else printf("%d\n",ans);}
}