小雀和他的王国【牛客练习赛56 E】【Tarjan缩点+树的直径】

题目链接

  首先，如果它本身就是在环内了，那么，任意的破坏环上的任意条边，都是不会影响答案的，所以，我们可以知道，会映像答案的边只有那些桥。

  于是，做法就变成了Tarjan缩点，然后就变成了一棵树了，我们现在想要构成最大的环，于是任务就变成了找最大的边构成一个环，那么问题再化简，就变成了求树的直径的问题了，于是我们直接推一下深度和高度，就可以来求树的直径了。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#define _ABS(x, y) ( x > y ? (x - y) : (y - x) )
#define lowbit(x) ( x&( -x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define efs 1e-7
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(a, b) make_pair(a, b)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const ll mod = 1e9 + 7;
ll fast_mi(ll a, ll b = mod - 2LL)
{
    ll ans = 1;
    while(b)
    {
        if(b & 1)
        {
            ans = ans * a % mod;
        }
        a = a * a % mod;
        b >>= 1;
    }
    return ans;
}
const int maxN = 2e5 + 7;
int N, M, head[maxN], cnt;
struct Eddge
{
    int nex, to, u;
    Eddge(int a=-1, int b=0, int c=0):nex(a), to(b), u(c) {}
}edge[maxN << 1];
inline void addEddge(int u, int v)
{
    edge[cnt] = Eddge(head[u], v, u);
    head[u] = cnt++;
}
inline void _add(int u, int v) { addEddge(u, v); addEddge(v, u); }
int dfn[maxN] = {0}, low[maxN], tot, Stap[maxN], Stop, Belong[maxN], Bcnt;
bool instack[maxN] = {false}, vis[maxN << 1] = {false};
void Tarjan(int u)
{
    dfn[u] = low[u] = ++tot;
    Stap[++Stop] = u;
    instack[u] = true;
    for(int i=head[u], v; ~i; i=edge[i].nex)
    {
        if(vis[i]) continue;
        vis[i] = vis[i ^ 1] = true;
        v = edge[i].to;
        if(!dfn[v])
        {
            Tarjan(v);
            low[u] = min(low[v], low[u]);
        }
        else if(instack[v]) low[u] = min(low[u], dfn[v]);
    }
    int v;
    if(low[u] == dfn[u])
    {
        Bcnt++;
        do
        {
            v = Stap[Stop--];
            instack[v] = false;
            Belong[v] = Bcnt;
        } while(u ^ v);
    }
}
int ans;
struct New_Graph
{
    int n, top[maxN], tp, du[maxN];
    struct Eg
    {
        int nex, to;
        Eg(int a=-1, int b=0):nex(a), to(b) {}
    }E[maxN << 1];
    inline void add_E(int u, int v)
    {
        E[tp] = Eg(top[u], v);
        top[u] = tp++;
    }
    inline void _E(int u, int v) { add_E(u, v); add_E(v, u); }
    int deep[maxN], high[maxN];
    void pre_dfs(int u, int fa)
    {
        int maxx = 0;
        for(int i=top[u], v; ~i; i=E[i].nex)
        {
            v = E[i].to;
            if(v == fa) continue;
            deep[v] = deep[u] + 1;
            pre_dfs(v, u);
            ans = max(ans, max(maxx + high[v], deep[u] + high[v]));
            if(high[v] > maxx)
            {
                maxx = high[v];
            }
        }
        high[u] = maxx + 1;
        ans = max(ans, deep[u]);
    }
    void Init()
    {
        n = Bcnt;
        for(int i=1; i<=n; i++)
        {
            top[i] = -1;
            du[i] = 0;
        }
        tp = 0;
        for(int i = 0, u, v; i < cnt; i += 2)
        {
            u = edge[i].u; v = edge[i].to;
            if(Belong[u] == Belong[v]) continue;
            _E(Belong[u], Belong[v]);
        }
        deep[1] = 0;
        pre_dfs(1, 0);
    }
}G;
inline void init()
{
    cnt = Stop = Bcnt = 0;
    for(int i=1; i<=N; i++) head[i] = -1;
}
int main()
{
    scanf("%d%d", &N, &M);
    init();
    for(int i=1, u, v; i<=M; i++)
    {
        scanf("%d%d", &u, &v);
        _add(u, v);
    }
    Tarjan(1);
    ans = 0;
    G.Init();
    printf("%lld\n", 1LL * (Bcnt - 1 - ans) * fast_mi(M + 1LL) % mod);
    return 0;
}