Codeforces 2127 D
D. Root was Built by Love, Broken by Destiny
题意: n栋房子，其中有m做桥分别连接两栋房子，然后把这些房子分别排列在南北两岸顺序不限，排列的结果不能有任意两座桥相交，求排列的方案数
思路: 把这个问题抽象成图论问题首先，然后由不相交的边联想到森林，则必须m = (n - 1)才存在合法方案数，否则一定会相交，假设把度数为1的节点去掉后，如果剩余是一条链则是合法的(特别注意如果一个点度数大于2的子节点个数大于2则不会构成一条链)，设链上的节点数为m则方案数为(\(\prod_{i=1}^m (cnt_i)!\))(cnt代表i节点度数为1的节点个数)。然后再考虑南北和链长度>=2时左右排列的两种状况。
Tag: 图论(抽象成树上问题，并最终抽象成一条链)，组合数学(非链上节点的排列及左右朝向)，DFS，分类讨论
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#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
#define endl "\n"
typedef pair<int, int> Pii;
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define REP3(i, m, n) for (int i = (m); (i) < int(n); ++ (i))
#define rep(i,b) for(int i=(int)(0);i<(int)(b);i++)
#define rng(i,a,b) for(int i=int(a);i<int(b);i++)
#define ALL(x) begin(x), end(x)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define fore(i,a) for(auto &i:a)
#define all(s) (s).begin(),(s).end()
#define drep2(i, m, n) for (int i = (m)-1; i >= (n); --i)
#define drep(i, n) drep2(i, n, 0)
#define rever(vec) reverse(vec.begin(), vec.end())
#define sor(vec) sort(vec.begin(), vec.end())
#define fi first
#define FOR_(n) for (ll _ = 0; (_) < (ll)(n); ++(_))
#define FOR(i, n) for (ll i = 0; (i) < (ll)(n); ++(i))
#define se second
#define pb push_back
#define P pair<ll,ll>
#define PQminll priority_queue<ll, vector<ll>, greater<ll>>
#define PQmaxll priority_queue<ll,vector<ll>,less<ll>>
#define PQminP priority_queue<P, vector<P>, greater<P>>
#define PQmaxP priority_queue<P,vector<P>,less<P>>
#define NP next_permutation
#define die(a) {cout<<a<<endl; return;}
#define dier(a) {return a;}
//const ll Mod = 1000000009;
//const ll Mod = 998244353;
const ll Mod = 1000000007;
#define inv(a) q_pow((ll)a, Mod - 2)
const ll inf = 4000000000000000000ll;
const ld eps = ld(0.00000000001);
static const long double pi = 3.141592653589793;
template<class T>void vcin(vector<T> &n){for(int i=0;i<int(n.size());i++) cin>>n[i];}
template<class T,class K>void vcin(vector<T> &n,vector<K> &m){for(int i=0;i<int(n.size());i++) cin>>n[i]>>m[i];}
template<class T>void vcout(vector<T> &n){for(int i=0;i<int(n.size());i++){cout<<n[i]<<" ";}cout<<endl;}
template<class T>void vcin(vector<vector<T>> &n){for(int i=0;i<int(n.size());i++){for(int j=0;j<int(n[i].size());j++){cin>>n[i][j];}}}
template<class T>void vcout(vector<vector<T>> &n){for(int i=0;i<int(n.size());i++){for(int j=0;j<int(n[i].size());j++){cout<<n[i][j]<<" ";}cout<<endl;}cout<<endl;}
void yes(bool a){cout<<(a?"yes":"no")<<endl;}
void YES(bool a){cout<<(a?"YES":"NO")<<endl;}
void Yes(bool a){cout<<(a?"Yes":"No")<<endl;}
void possible(bool a){ cout<<(a?"possible":"impossible")<<endl; }
void Possible(bool a){ cout<<(a?"Possible":"Impossible")<<endl; }
void POSSIBLE(bool a){ cout<<(a?"POSSIBLE":"IMPOSSIBLE")<<endl; }
#define FOR_R(i, n) for (ll i = (ll)(n)-1; (i) >= 0; --(i))
template<class T>auto min(const T& a){ return *min_element(all(a)); }
//template<class T>auto max(const T& a){ return *max_element(all(a)); }
template<class T,class F>void print(pair<T,F> a){cout<<a.fi<<" "<<a.se<<endl;}
template<class T>bool chmax(T &a,const T b) { if (a<b) { a=b; return 1; } return 0;}
template<class T>bool chmin(T &a,const T b) { if (b<a) { a=b; return 1; } return 0;}
template<class T> void ifmin(T t,T u){if(t>u){cout<<-1<<endl;}else{cout<<t<<endl;}}
template<class T> void ifmax(T t,T u){if(t>u){cout<<-1<<endl;}else{cout<<t<<endl;}}
ll fastgcd(ll u,ll v){ll shl=0;while(u&&v&&u!=v){bool eu=!(u&1);bool ev=!(v&1);if(eu&&ev){++shl;u>>=1;v>>=1;}else if(eu&&!ev){u>>=1;}else if(!eu&&ev){v>>=1;}else if(u>=v){u=(u-v)>>1;}else{ll tmp=u;u=(v-u)>>1;v=tmp;}}return !u?v<<shl:u<<shl;}
ll modPow(ll a, ll n, ll Mod) { if(Mod==1) return 0;ll ret = 1; ll p = a % Mod; while (n) { if (n & 1) ret = ret * p % Mod; p = p * p % Mod; n >>= 1; } return ret; }
vector<ll> divisor(ll x){ vector<ll> ans; for(ll i = 1; i * i <= x; i++){ if(x % i == 0) {ans.push_back(i); if(i*i!=x){ ans.push_back(x / ans[i]);}}}sor(ans); return ans; }
ll pop(ll x){return __builtin_popcountll(x);}
ll poplong(ll x){ll y=-1;while(x){x/=2;y++;}return y;}
P hyou(P a){ll x=fastgcd(abs(a.fi),abs(a.se));a.fi/=x;a.se/=x;if(a.se<0){a.fi*=-1;a.se*=-1;}return a;}
P Pplus(P a,P b){ return hyou({a.fi*b.se+b.fi*a.se,a.se*b.se});}
P Ptimes(P a,ll b){ return hyou({a.fi*b,a.se});}
P Ptimes(P a,P b){ return hyou({a.fi*b.fi,a.se*b.se});}
P Pminus(P a,P b){ return hyou({a.fi*b.se-b.fi*a.se,a.se*b.se});}
P Pgyaku(P a){ return hyou({a.se,a.fi});}ll q_pow(ll x, ll n) {ll res = 1;for (; n > 0; n /= 2) {if (n % 2) res = res * x % Mod;x = x * x % Mod;}return res % Mod;
}ll i, j;
ll n, m, k;
ll l, r, p, q;
ll a, b;
ll mul[200005];
void solve() {cin >> n >> m;vector<vector<int>> g(n);vector<int> cnt(n);for(int i = 0; i < m; ++i) {cin >> a >> b;--a;--b;g[a].pb(b);g[b].pb(a);cnt[a]++;cnt[b]++;} if(m != (n - 1)) {cout << 0 << endl;return;} ll count = 0;vector<bool> flag(n, false);for(int i = 0; i < n; ++i) {int nxt = 0;if(cnt[i] == 1) continue;for(auto j : g[i]) {if(cnt[j] >= 2) {if(nxt >= 2) { // 如果度数为2的点过多，则不能成链，方案数为0cout << 0 << endl;return;}else {nxt++;} // 记录度数大于等于2的节点数}}flag[i] = true; // 处于链子上count++;}ll res = 2LL; // 南北 for(int i = 0; i < n; ++i) {if(!flag[i]) continue;ll c = 0;for(auto j : g[i]) { // 记录链上度数节点为1的数量if(cnt[j] == 1 && !flag[j]) {c++;} }res = (res * mul[c]) % Mod; // 累乘链上度数为1的节点数量的阶乘，等价于这些节点的任意排列}if(count >= 2) res = (res * 2LL) % Mod; // 如果链子的长度大于2，则需要考虑链子左右的排列cout << res << endl;
}
int main() {std::ios::sync_with_stdio(false);std::cin.tie(0);std::cout.tie(0);mul[0] = 1;for(ll i = 1; i < 200005; ++i) {mul[i] = mul[i - 1] * i;mul[i] %= Mod;}int t;cin >> t;while(t--) solve();
}