加一条边后最少还有多少个桥，先Tarjan双联通缩点，
然后建树，求出树的直径，在直径起点终点加一条边去的桥最多，
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stack>
#define N 200001
using namespace std;
int belong[N],head[N],num,ins[N],n,dfs[N],low[N],idx,ans,num1;
struct edge
{
int st,ed,next;
}E[N*10],e[N*10];
void addedge(int x,int y)
{
E[num].st=x;
E[num].ed=y;
E[num].next=head[x];
head[x]=num++;
}
void Addedge(int x,int y)
{
e[num1].st=x;
e[num1].ed=y;
e[num1].next=head[x];
head[x]=num1++;
}
stack<int>Q;
void Tarjan(int u,int father)
{
int i,j,v,flag=0;
low[u]=dfs[u]=idx++;
Q.push(u);
for(i=head[u];i!=-1;i=E[i].next)
{
j=E[i].ed;
if(dfs[j]==-1)
{
Tarjan(j,u);
low[u]=low[u]>low[j]?low[j]:low[u];
}
else if(j==father)
{
if(flag)
low[u]=low[u]>dfs[j]?dfs[j]:low[u];
flag++; }
else low[u]=low[u]>dfs[j]?dfs[j]:low[u]; }
if(dfs[u]==low[u])
{
do
{
v=Q.top();
Q.pop();
belong[v]=ans;
}while(v!=u);
ans++;
}
}
void read()
{
memset(head,-1,sizeof(head));
num1=0;
for(int i=0;i<num;i+=2)
{
int x=belong[E[i].st];
int y=belong[E[i].ed];
if(x==y)continue;
Addedge(x,y);
Addedge(y,x);
}
}
int dis;
int dist(int u)
{
ins[u]=1;
int max=0,mmax=0;
for(int i=head[u];i!=-1;i=e[i].next)
{
int v=e[i].ed;
if(ins[v]==1)continue;
int temp=dist(v);
if(temp>mmax)
{
max=mmax;
mmax=temp;
}
else if(temp>max)
{
max=temp;
}
}
if(dis<(mmax+max+1))
dis=mmax+max+1;
return mmax+1;
}
int main()
{
int i,m,x,y;
while(scanf("%d%d",&n,&m),n||m)
{
memset(head,-1,sizeof(head));
num=0;
for(i=0;i<m;i++)
{
scanf("%d%d",&x,&y);
addedge(x,y);
addedge(y,x);
}
memset(dfs,-1,sizeof(dfs));
memset(low,-1,sizeof(low));
memset(ins,0,sizeof(ins));
idx=ans=0;
Tarjan(1,-1);
read();
memset(ins,0,sizeof(ins));
dis=0;
dist(1);
printf("%d\n",ans-dis);
}
return 0;
}