F

题意：有一个无向连通图，现在问添加一条边后最少还有几个桥

分析：先把图缩点，然后重构图为一棵树，求出来树的直径即可，不过注意会有重边，构树的时候注意一下

***********************************************************************

#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;

const int MAXN = 2e5+5;

struct Edge{int v, vis, next;}e1[MAXN*10], e2[MAXN*10];
int Head1[MAXN], Head2[MAXN], cnt[3], fa[MAXN];
void AddEdge(Edge e[], int Head[], int u, int v, int k)
{
    e[ cnt[k] ].v = v;
    e[ cnt[k] ].vis = 0;
    e[ cnt[k] ].next = Head[u];
    Head[u] = cnt[k]++;
}

struct node{int u, step;};
node BFS(node s, int k)
{
    queue<node> Q;
    Q.push(s);

    while(Q.size())
    {
        s = Q.front();Q.pop();

        for(int j=Head2[s.u]; j!=-1; j=e2[j].next)
        {
            node q = s;
            q.u = e2[j].v;

            if(e2[j].vis != k)
            {
                e2[j].vis = e2[j^1].vis = k;
                q.step++;
                Q.push(q);
            }
        }

    }

    return s;
}

int dfn[MAXN], low[MAXN], Index;
int belong[MAXN], bnt;
int Stack[MAXN], top;

void InIt(int N)
{
    cnt[1] = cnt[2] = Index = bnt = top = 0;
    for(int i=1; i<=N; i++)
    {
        Head1[i] = Head2[i] = -1;
        dfn[i] = 0;
        fa[i] = 0;
    }
}
void Tarjan(int u)
{
    int v;

    low[u] = dfn[u] = ++Index;
    Stack[++top] = u;

    for(int j=Head1[u]; j!=-1; j=e1[j].next)
    {
        v = e1[j].v;
        if(e1[j].vis == false)
        {
            e1[j].vis = e1[j^1].vis = true;
            if( !dfn[v] )
            {
                Tarjan(v);
                low[u] = min(low[u], low[v]);
            }
            else
                low[u] = min(low[u], dfn[v]);
        }
    }

    if(low[u] == dfn[u])
    {
        ++bnt;
        do
        {
            v = Stack[top--];
            belong[v] = bnt;
        }
        while(u != v);
    }
}
int main()
{
    int N, M;

    while(scanf("%d%d", &N, &M), N+M)
    {
        int i, j, u, v;

        InIt(N);

        while(M--)
        {
            scanf("%d%d", &u, &v);
            AddEdge(e1, Head1, u, v, 1);
            AddEdge(e1, Head1, v, u, 1);
        }

        Tarjan(1);

        for(i=1; i<=N; i++)
        for(j=Head1[i]; j!=-1; j=e1[j].next)
        {
            v = e1[j].v;
            u = belong[i], v = belong[v];
            if(u > v && fa[v] != u)
            {
                fa[v] = u;
                AddEdge(e2, Head2, u, v, 2);
                AddEdge(e2, Head2, v, u, 2);
            }
        }

        node s;
        s.u = 1, s.step = 0;

        s = BFS(s, 1);
        s.step = 0;
        s = BFS(s, 2);

        printf("%d
", bnt-s.step-1);
    }

    return 0; 

}