（求树的直径）Warm up -- HDU -- 4612

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
#define N 200005
int Head1[N], Head2[N], cnt[3];
int Stack[N], top, dfn[N], low[N], Time, n, m;
int nBridge, Bridge[N];
int dist[N], vis[N], Max, index;
struct Edge
{
    int v, next;
} e1[10*N], e2[10*N];
void Init()
{
    top = nBridge = Time = Max = index = 0;
    cnt[1] = cnt[0] = 0;
    memset(low, 0, sizeof(low));
    memset(dfn, 0, sizeof(dfn));
    memset(Bridge, 0, sizeof(Bridge));
    memset(dist, 0, sizeof(dist));
    memset(Stack, 0, sizeof(Stack));
    memset(Head1, -1, sizeof(Head1));
    memset(Head2, -1, sizeof(Head2));
}
void Add(Edge e[],int Head[], int u, int v, int k)
{
    e[cnt[k]].v = v;
    e[cnt[k]].next = Head[u];
    Head[u] = cnt[k]++;
}
void Tarjar(int u, int father)
{
    low[u] = dfn[u] = ++Time;
    Stack[top++] = u;
    int v, k=0;
    for(int i=Head1[u]; i!=-1; i=e1[i].next)
    {
        v = e1[i].v;
        if(v==father && !k)///避免重边；
        {
            k++;
            continue;
        }
        if(!dfn[v])
        {
            Tarjar(v, u);
            low[u] = min(low[u], low[v]);
        }
        else
            low[u] = min(low[u], dfn[v]);
    }
    if(low[u] == dfn[u])
    {
        nBridge++;///可以代表缩点后的节点个数；
        while(1)
        {
            v = Stack[--top];
            Bridge[v] = nBridge;///缩点；
            if(u==v)
                break;
        }
    }
}
void bfs(int s)
{
    queue<int>Q;
    int p, q;
    memset(vis, 0, sizeof(vis));
    vis[s] = 1;
    dist[s] = 0;
    Q.push(s);
    while(!Q.empty())
    {
        p = Q.front(); Q.pop();
        for(int i=Head2[p]; i!=-1; i=e2[i].next)
        {
            q = e2[i].v;
            if(!vis[q])
            {
                vis[q] = 1;
                dist[q] = dist[p] + 1;
                Q.push(q);
                if(Max<dist[q])
                {
                    Max = dist[q];
                    index = q;
                }
            }
        }
    }
}
int main()
{
    int u, v;
    while(scanf("%d%d", &n, &m), m + n)///输入时由于m n弄反了，TLE的我想哭；
    {
        Init();
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d", &u, &v);
            Add(e1, Head1, u, v, 0);
            Add(e1, Head1, v, u, 0);///原来的树；
        }
        Tarjar(1, 0);
        for(int i=1; i<=n; i++)
        {
            for(int j=Head1[i]; j!=-1; j=e1[j].next)
            {
                int u = Bridge[i];
                int v = Bridge[e1[j].v];
                if(u != v )
                {
                    Add(e2, Head2, u, v, 1);
                    Add(e2, Head2, v, u, 1);///缩点后的树；
                }
            }
        }
        bfs(1);
        bfs(index);///求树的直径的过程；
        printf("%d\n", nBridge-1-Max);///缩点后形成的树每条边都是桥；所以总桥的个数为节点数-1；
    }
    return 0;
}