將未建立貿易關係看成連一條邊，那麼這顯然是個二分圖。最大城市群即最大獨立集，也即n-最大匹配。現在要求的就是刪哪些邊會使最大匹配減少，也即求哪些邊一定在最大匹配中。

　　首先範圍有點大，當然是跑個dinic，轉化成最大流。會使最大流減少的邊相當於可能在最小割中的邊，因為刪掉它就相當於無代價的割掉了一條邊。那麼用曾經看到過的結論就可以了：當且僅當該邊滿流且殘餘網路（包括反向邊）中該邊兩端點處於不同SCC時，該邊可能在最小割中。不太會證。於是tarjan一發就可以了。注意不要把開始給的圖和網路流建圖搞混。

#include<iostream> 
#include<cstdio>
#include
 
<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 10010
#define M 300010
#define S 0
#define T 10001
char getc(){char c=getchar();while (c==10||c==13||c==32) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
 
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,m,p[N],t=-1,color[N],ans;
int d[N],q[N],cur[N];
struct data{int to,nxt,cap,flow;
}edge[M
 
<<1];
struct data2
{
    int x,y;
    bool operator <(const data2&a) const
    {
        return x<a.x||x==a.x&&y<a.y;
    }
}v[M];
void addedge(int x,int y,int z)
{
    t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].cap=z,edge[t].flow=0,p[x]=t;
    t++;edge[t].to=x,edge[t].nxt=p[y],edge[t].cap=0,edge[t].flow=0,p[y]=t;
}
void paint(int k)
{
    for (int i=p[k];~i;i=edge[i].nxt)
    if (color[edge[i].to]==-1)
    {
        color[edge[i].to]=color[k]^1;
        paint(edge[i].to);
    }
}
bool bfs()
{
    int head=0,tail=1;q[1]=S;
    memset(d,255,sizeof(d));d[S]=0;
    do
    {
        int x=q[++head];
        for (int i=p[x];~i;i=edge[i].nxt)
        if (d[edge[i].to]==-1&&edge[i].flow<edge[i].cap)
        {
            d[edge[i].to]=d[x]+1;
            q[++tail]=edge[i].to;
        }
    }while (head<tail);
    return ~d[T];
}
int work(int k,int f)
{
    if (k==T) return f;
    int used=0;
    for (int i=cur[k];~i;i=edge[i].nxt)
    if (d[k]+1==d[edge[i].to])
    {
        int w=work(edge[i].to,min(f-used,edge[i].cap-edge[i].flow));
        edge[i].flow+=w,edge[i^1].flow-=w;
        if (edge[i].flow<edge[i].cap) cur[k]=i;
        used+=w;if (used==f) return f;
    }
    if (used==0) d[k]=-1;
    return used;
}
void dinic()
{
    while (bfs())
    {
        memcpy(cur,p,sizeof(p));
        work(S,N);
    }
}
namespace newgraph
{
    int dfn[N]={0},low[N]={0},stk[N],id[N],top=0,cnt=0,tot=0,t=0,p[N]={0},ans=0;
    bool flag[N];
    struct data{int to,nxt;}edge[M];
    void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}
    void tarjan(int k)
    {
        dfn[k]=low[k]=++cnt;
        stk[++top]=k;flag[k]=1;
        for (int i=p[k];i;i=edge[i].nxt)
        if (!dfn[edge[i].to]) tarjan(edge[i].to),low[k]=min(low[k],low[edge[i].to]);
        else if (flag[edge[i].to]) low[k]=min(low[k],dfn[edge[i].to]);
        if (dfn[k]==low[k])
        {
            tot++;
            while (stk[top]!=k)
            {
                flag[stk[top]]=0;
                id[stk[top]]=tot;
                top--;
            }
            flag[k]=0;id[k]=tot;top--;
        }
    }
    void work()
    {
        for (int i=0;i<=n;i++)
        if (!dfn[i]) tarjan(i);
        if (!dfn[T]) tarjan(T);
    }
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read();
    memset(p,255,sizeof(p));
    for (int i=1;i<=m;i++)
    {
        int x=read(),y=read();
        addedge(x,y,1);
    }
    memset(color,255,sizeof(color));
    for (int i=1;i<=n;i++) if (color[i]==-1) color[i]=1,paint(i);
    for (int i=0;i<=t;i++) edge[i].cap=color[edge[i^1].to];
    for (int i=1;i<=n;i++)
    if (color[i]) addedge(S,i,1);
    else addedge(i,T,1);
    dinic();
    for (int i=0;i<=t;i++)
    if (edge[i].flow<edge[i].cap) newgraph::addedge(edge[i^1].to,edge[i].to);
    newgraph::work();
    for (int i=0;i<=t;i++)
    if (edge[i].cap==1&&edge[i].flow==edge[i].cap&&edge[i^1].to!=S&&edge[i].to!=T&&newgraph::id[edge[i^1].to]!=newgraph::id[edge[i].to])
    ans++,v[ans].x=min(edge[i^1].to,edge[i].to),v[ans].y=max(edge[i^1].to,edge[i].to);
    sort(v+1,v+ans+1);
    cout<<ans<<endl;
    for (int i=1;i<=ans;i++) printf("%d %d\n",v[i].x,v[i].y);
    return 0;
}