UVA11174 Stand in a Line

题意

PDF

分析

\[
f(i)=f(c_1)f(c_2)\dots\times(s(i)-1)!/(s(c_1)!s(c_2)! \dots s(c_k)! )\\
f(root)=(s(root)-1)!/(s(1)s(2)s(3)\dots s(n))
\]
预处理阶乘以及逆元。

时间复杂度\(O(Tn)\)

代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<ctime>
#include<cstring>
#define rg register
#define il inline
#define co const
template<class T>il T read()
{
    rg T data=0;
    rg int w=1;
    rg char ch=getchar();
    while(!isdigit(ch))
    {
        if(ch=='-')
            w=-1;
        ch=getchar();
    }
    while(isdigit(ch))
    {
        data=data*10+ch-'0';
        ch=getchar();
    }
    return data*w;
}
template<class T>T read(T&x)
{
    return x=read<T>();
}
using namespace std;
typedef long long ll;
co int N=4e4+1,mod=1e9+7;
int fac[N],inv[N];
int mul(int x,int y)
{
    return (ll)x*y%mod;
}
int qpow(int x,int k)
{
    int res=1;
    while(k)
    {
        if(k&1)
            res=mul(res,x);
        x=mul(x,x),k>>=1;
    }
    return res;
}
int n,fa[N];
int nx[N],to[N];
int siz[N];
void dfs(int x)
{
    siz[x]=1;
    for(int i=to[x];i;i=nx[i])
    {
        dfs(i);
        siz[x]+=siz[i];
    }
}
int main()
{
//  freopen(".in","r",stdin);
//  freopen(".out","w",stdout);
    fac[0]=fac[1]=1;
    inv[0]=inv[1]=1;
    for(int i=2;i<N;++i)
    {
        fac[i]=mul(i,fac[i-1]);
        inv[i]=mul(mod-mod/i,inv[mod%i]);
    }
    int T=read<int>();
    while(T--)
    {
        read(n);
        fill(fa,fa+n+1,0);
        fill(nx,nx+n+1,0);
        fill(to,to+n+1,0);
        int m=read<int>();
        for(int i=1;i<=m;++i)
        {
            int a=read<int>(),b=read<int>();
            fa[a]=b;
            nx[a]=to[b],to[b]=a;
        }
        for(int i=1;i<=n;++i)
            if(!fa[i])
                nx[i]=to[0],to[0]=i;
        dfs(0);
        int ans=fac[n];
        for(int i=1;i<=n;++i)
            ans=mul(ans,inv[siz[i]]);
        printf("%d\n",ans);
    }
    return 0;
}