URAL 1962 In Chinese Restaurant 数学

In Chinese Restaurant

题目连接：

http://acm.hust.edu.cn/vjudge/contest/123332#problem/B

Description

When Vova arrived in Guangzhou, his Chinese friends immediately invited him to a restaurant. Overall n people came to the restaurant, including Vova. The waiter offered to seat the whole company at a traditional large round table with a rotating dish plate in the center.
As Vova was a guest, he got the honorable place by the door. Then m people in the company stated that they wanted to sit near a certain person. Your task is to determine the number of available arrangements to seat Vova's friends at the table.

Input

The first line contains integers n and m (2 ≤ n ≤ 100; 0 ≤ m ≤ n). The next m lines contain integers k1, …, km, where ki is the number of the person who the person number i wants to sit with (1 ≤ ki ≤ n; ki ≠ i). Being an honorable guest, Vova got number 1. His friends are numbered with integers from 2 to n.

Output

Print the number of available arrangements to seat Vova's friends modulo 10 9 + 7.

Sample Input

6 6
2
1
1
5
6
5

Sample Output

4

Hint

题意

有n个人在坐在一个圆桌上，给你m个关系a,b,表示a要挨着b坐

问你有多少个方案数

题解：

如果出现环，答案为2，否则就是全排列乘以2

特判掉 只有两个人的情况，因为这样正着坐和反着坐答案是一样的。

代码

#include <bits/stdc++.h>
#define rep(a,b,c) for(int (a)=(b);(a)<=(c);++(a))
#define drep(a,b,c) for(int (a)=(b);(a)>=(c);--(a))
#define pb push_back
#define mp make_pair
#define sf scanf
#define pf printf
#define two(x) (1<<(x))
#define clr(x,y) memset((x),(y),sizeof((x)))
#define dbg(x) cout << #x << "=" << x << endl;
const int mod = 1e9 + 7;
int mul(int x,int y){return 1LL*x*y%mod;}
int qpow(int x , int y){int res=1;while(y){if(y&1) res=mul(res,x) ; y>>=1 ; x=mul(x,x);} return res;}
inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;}
using namespace std;

const int maxn = 1e3 + 15;
vector<int>G[maxn],block[maxn];
int n , m ,vis[maxn] , c[maxn],tot,sz[maxn],deg[maxn],fac[maxn],C[maxn][maxn],moo[maxn][maxn];
set < int > Edge[maxn];

void maintain(int & x){
    if( x >= mod ) x-= mod;
}

bool dfs1( int x , int pre ){
	c[x] = 1;
	for(auto v:G[x]){
		if( v == pre ) continue;
		if(c[v] == 1) return true;
		else if(c[v] == 0){
			if(dfs1(v,x)) return true;
		}
	}
	c[x] = 2;
	return false;
}

void predfs( int x ){
	block[tot].pb( x );
	for(auto v : G[x])
		if( vis[v] == 0 ){
			vis[v] = 1;
			predfs( v );
		}
}

void init(){
	fac[0] = 1;
	C[0][0] = 1;
	rep(i,1,maxn-1){
		fac[i] = mul( fac[i - 1] , i );
		C[i][0] = 1;
		rep(j,1,i){
			C[i][j] = C[i][j - 1] + C[i - 1][j - 1] ;
			maintain( C[i][j] );
		}
	}
}

int main(int argc,char *argv[]){
	init();
	n=read(),m=read();
	if(n == 2 && m >= 1 ){
		pf("1
");
		return 0;
	}
	rep(i,1,n) G[i].clear();
	rep(i,1,m){
		int v=read();
		Edge[i].insert( v );
		moo[i][v]=moo[v][i]=1;
	}
	rep(i,1,n){
		rep(j,1,n) if( moo[i][j]  && i != j ) G[i].pb( j );
	}
	int ok = 0;
	rep(i,1,n) if(c[i] == 0){
		if( dfs1( i , 0 ) ){
			ok = 1;
			break;
		}
	}
	rep(i,1,n) if(!vis[i]){
		vis[i]=1;
		++ tot;
		predfs( i );
	}
	// 有三条边或以上
	rep(i,1,n){
		int kk = 0;
		rep(j,1,n) if( moo[i][j]) ++ kk;
		if( kk > 2 ){
			pf("0
");
			return 0;
		}
	}
	if( ok ){
		if( tot == 1 ){
			int judge = 1;
			rep(i,1,n) if(c[i] != 1) judge = 0;
			if( judge == 1 ) pf("%d
" , 2);
			else pf("0
");
		}else pf("0
");
		return 0;
	}else{
		int ptr = 0 ;
		rep(i,1,tot){
			for(auto it : block[i]){
				if( it == 1 ){
					ptr = i;
				}
			}
		}
		int ans = 1 ;
		if( block[ptr].size() > 1 ) ans = 2;
		rep(i,1,tot){
			if( i == ptr ) continue;
			int cap = block[i].size();
			if( cap > 1 ) ans = mul( ans , 2 );
		}
		ans = mul( ans , fac[ tot - 1 ] );
		pf("%d
" , ans );
	}
	return 0;
}