hdu 4612 Warm up(邊-雙連通+縮點+樹的直徑)

題意是給定一個無向圖，求增加一條邊後，橋的最少可能的條數。

先求出所有橋（即雙連通分量），然後縮點得到一顆樹。增加一條邊使得橋的數量最小，顯然是連接bcc樹上直徑的兩端了。

這個題hdu又會爆棧。。。開個掛才能過。。。

#pragma comment(linker,"/STACK:102400000,102400000")
#include<iostream>
#include<algorithm>
#include<vector>
#include<string>
#include<queue>
#include<stack>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<fstream>
#include<sstream>
#include<map>
#include<set>
#define FF(i, a, b) for(int i=a; i<b; i++)
#define FD(i, a, b) for(int i=a; i>=b; i--)
#define REP(i, n) for(int i=0; i<n; i++)
#define CLR(a, b) memset(a, b, sizeof(a))
#define LL long long
#define PB push_back
#define debug puts("**debug**")
using namespace std;

const int maxn = 200001;
int n, m, u, v;
int pre[maxn], low[maxn], dfs_clock, bcc_cnt, bccno[maxn];
struct Edge
{
    int to, flag;
};
vector<int> G[maxn], g[maxn];
vector<Edge> edges;

inline void init()
{
    REP(i, n) G[i].clear(); edges.clear();
}

void add(int u, int v)
{
    Edge e; e.to = v, e.flag = 0;
    edges.PB(e);
    e.to = u;
    edges.PB(e);
    int nc = edges.size();
    G[u].PB(nc-2);
    G[v].PB(nc-1);
}

int dfs(int u, int fa)
{
    int lowu = pre[u] = ++dfs_clock;
    int nc = G[u].size();
    REP(i, nc)
    {
        int v = edges[G[u][i]].to;
        if(!pre[v])
        {
            int lowv = dfs(v, u);
            lowu = min(lowu, lowv);
            if(lowv > pre[u]) edges[G[u][i]].flag = 1, edges[G[u][i]^1].flag = 1;
        }
        else if(pre[v] < pre[u] && v != fa) lowu = min(lowu, pre[v]);
    }
    return low[u] = lowu;
}

void dfs1(int u)
{
    bccno[u] = bcc_cnt;
    int nc = G[u].size();
    REP(i, nc)
    {
        int v = edges[G[u][i]].to;
        if(!bccno[v] && edges[G[u][i]].flag != 1) dfs1(v);
    }
}

void find_bcc()
{
    CLR(pre, 0); CLR(bccno, 0);
    dfs_clock = bcc_cnt = 0;
    REP(i, n) if(!pre[i]) dfs(i, -1);
    REP(i, n) if(!bccno[i]) bcc_cnt++, dfs1(i);
}

int mm, end;

void find(int u, int fa, int d)
{
    int nc = g[u].size();
    if(nc == 1)
    {
        if(d > mm)
        {
            mm = d;
            end = u;
        }
    }
    REP(i, nc)
    {
        int v = g[u][i];
        if(v != fa) find(v, u, d+1);
    }
}

int main()
{
    while(scanf("%d%d", &n, &m), n+m)
    {
        init();
        REP(i, m)
        {
            scanf("%d%d", &u, &v); u--; v--;
            add(u, v);
        }
        find_bcc();
        if(bcc_cnt == 1)
        {
            puts("0");
            continue;
        }
        REP(i, n+1) g[i].clear();
        REP(u, n)
        {
            int nc = G[u].size();
            REP(i, nc)
            {
                int v = edges[G[u][i]].to;
                if(bccno[u] != bccno[v]) g[bccno[u]].PB(bccno[v]);
            }
        }
        mm = 0;
        find(1, -1, 0);
        find(end, -1, 0);
        printf("%d\n", bcc_cnt - mm - 1);
    }
    return 0;
}