UVA11174 Stand in a Line

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]])

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
inline void swap(long long &a, long long &b)
{
long long tmp = a;a = b;b = tmp;
}
inline void read(long long &x)
{
x = 0;char ch = getchar(), c = ch;
while(ch < '0' || ch > '9') c = ch, ch = getchar();
while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
if(c == '-') x = -x;
}

const long long INF = 0x3f3f3f3f;
const long long MAXN = 100000 + 10;
const long long MOD = 1000000007;

struct Edge
{
long long u,v,nxt;
Edge(long long _u, long long _v, long long _nxt){u = _u;v = _v;nxt = _nxt;}
Edge(){}
}edge[MAXN << 1];
long long head[MAXN], cnt;
inline void insert(long long a, long long b)
{
edge[++cnt] = Edge(a,b,head[a]);
head[a] = cnt;
}

long long f[MAXN],t,n,m,b[MAXN],size[MAXN],fa[MAXN];

void dfs(long long u)
{
size[u] = 1;b[u] = 1;
for(register long long pos = head[u];pos;pos = edge[pos].nxt)
{
long long v = edge[pos].v;
if(b[v]) continue;
dfs(v), size[u] += size[v];
}
return;
}

long long pow(long long a, long long b)
{
long long r = 1, base = a%MOD;
for(;b;b >>= 1)
{
if(b & 1) r *= base, r %= MOD;
base *= base, base %= MOD;
}
return r;
}

long long ni(long long x)
{
return pow(x, MOD - 2);
}

int main()
{
read(t);
f[0] = 1;
for(register long long i = 1;i < MAXN;++ i)
f[i] = (f[i - 1] * i) % MOD;
for(;t;--t)
{
memset(fa, 0, sizeof(fa));memset(head, 0, sizeof(head)), memset(b, 0, sizeof(b)), cnt = 0;
read(n), read(m);
for(register long long i = 1;i <= m;++ i)
{
long long tmp1,tmp2;
read(tmp1), read(tmp2);
insert(tmp2, tmp1);
fa[tmp1] = tmp2;
}
for(register long long i = 1;i <= n;++ i)
if(!b[i])
{
int now = i;
while(fa[now]) now = fa[now];
dfs(now);
}
long long ans = 1;
for(register long long i = 1;i <= n;++ i) ans *= size[i], ans %= MOD;
printf("%lld\n", f
* ni(ans) % MOD);
}
return 0;
}

UVA11174