Stand in a Line
大意：有n个人排队，问有多少种排列使得没有人排在他的父亲前面
不难发现这是一个森林
设一个虚根root把所有树的根连起来，root排在所有方案的最前面，总方案数不变
设i的儿子为son1(i),son2(i)....sonk(i),k位i儿子的数量
设size[i]为i这棵子树的节点个数
f[i]为i这颗子树的方案数
f[i] = f[son1[i]] * f[son2[i]] * ... * f[sonk[i]] * (size[i] - 1) / (size[son1[i]]! * size[son2[i]]! * .. * size[sonk[i]]!)
相当于每颗子树内部排列：f[son1[i]] * f[son2[i]] * ... * f[sonk[i]]
然后把同一颗子树看做相同的排：(size[i] - 1) / (size[son1[i]]! * size[son2[i]]! * .. * size[sonk[i]]!)
把f[son1[i]]...f[sonk[i]拆开，带进去
发现size[i]！作为分母出现一次，（size[i]-1）!作为分母出现一次，约分得size[i]
于是f[root] = (size[root] - 1)!/(size[son1[i]] * size[son2[i]] * ... * size[sonk[i]])
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <algorithm> 6 #include <queue> 7 #include <vector> 8 #include <cmath> 9 #define min(a, b) ((a) < (b) ? (a) : (b)) 10 #define max(a, b) ((a) > (b) ? (a) : (b)) 11 #define abs(a) ((a) < 0 ? (-1 * (a)) : (a)) 12 inline void swap(long long &a, long long &b) 13 { 14 long long tmp = a;a = b;b = tmp; 15 } 16 inline void read(long long &x) 17 { 18 x = 0;char ch = getchar(), c = ch; 19 while(ch < '0' || ch > '9') c = ch, ch = getchar(); 20 while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar(); 21 if(c == '-') x = -x; 22 } 23 24 const long long INF = 0x3f3f3f3f; 25 const long long MAXN = 100000 + 10; 26 const long long MOD = 1000000007; 27 28 struct Edge 29 { 30 long long u,v,nxt; 31 Edge(long long _u, long long _v, long long _nxt){u = _u;v = _v;nxt = _nxt;} 32 Edge(){} 33 }edge[MAXN << 1]; 34 long long head[MAXN], cnt; 35 inline void insert(long long a, long long b) 36 { 37 edge[++cnt] = Edge(a,b,head[a]); 38 head[a] = cnt; 39 } 40 41 long long f[MAXN],t,n,m,b[MAXN],size[MAXN],fa[MAXN]; 42 43 void dfs(long long u) 44 { 45 size[u] = 1;b[u] = 1; 46 for(register long long pos = head[u];pos;pos = edge[pos].nxt) 47 { 48 long long v = edge[pos].v; 49 if(b[v]) continue; 50 dfs(v), size[u] += size[v]; 51 } 52 return; 53 } 54 55 long long pow(long long a, long long b) 56 { 57 long long r = 1, base = a%MOD; 58 for(;b;b >>= 1) 59 { 60 if(b & 1) r *= base, r %= MOD; 61 base *= base, base %= MOD; 62 } 63 return r; 64 } 65 66 long long ni(long long x) 67 { 68 return pow(x, MOD - 2); 69 } 70 71 int main() 72 { 73 read(t); 74 f[0] = 1; 75 for(register long long i = 1;i < MAXN;++ i) 76 f[i] = (f[i - 1] * i) % MOD; 77 for(;t;--t) 78 { 79 memset(fa, 0, sizeof(fa));memset(head, 0, sizeof(head)), memset(b, 0, sizeof(b)), cnt = 0; 80 read(n), read(m); 81 for(register long long i = 1;i <= m;++ i) 82 { 83 long long tmp1,tmp2; 84 read(tmp1), read(tmp2); 85 insert(tmp2, tmp1); 86 fa[tmp1] = tmp2; 87 } 88 for(register long long i = 1;i <= n;++ i) 89 if(!b[i]) 90 { 91 int now = i; 92 while(fa[now]) now = fa[now]; 93 dfs(now); 94 } 95 long long ans = 1; 96 for(register long long i = 1;i <= n;++ i) ans *= size[i], ans %= MOD; 97 printf("%lld ", f[n] * ni(ans) % MOD); 98 } 99 return 0; 100 }