URAL1962:In Chinese Restaurant(并查集)

Open Ural FU Personal Contest 2013

When Vova arrived in Guangzhou, his Chinese friends immediately invited him to a restaurant. Overalln 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 109 + 7.

Samples

inputoutput

6 6211565

4

4 3231

0

题意：有n个人，分别编号1~n，有m个人分别表达自己的意愿，这m个人分别是1~m编号的这几个人

然后给出m个数，表示第i个人想要坐在第x个人的旁边，现在要求的是，在满足了这m个人的意愿的情况下，这张大圆桌有几种坐法

思路：我们首先来分析，对于那些想要坐在临近的人必然是互相相连的，那么他们就可以构成一棵树，那么对于总共的坐法，就是对s棵树的全排列

对于单个点的树，我们可以不用太多考虑

那么对于有子节点的树，我们要考虑环的问题

如果成环了，那么必然只剩这棵树本身，如果还有其他子树，那么必然无法满足

对于不成环的情况，我们还要注意的是对于单个节点，如果其身边必须安排的人数大于2，那么是无法满足的，因为他身边最多只能安排2个人

那么对于每棵树，这棵树不只一个节点，那么必然有两种情况

首先，如果是一条直链，例如1-2，我们有1-2和2-1两种

对于不是直链，那么肯定可以交换左右两支

什么？你说三链？这种情况之前就否定了，是0

解决了这些问题之后，这道题就解决了

#include <iostream>#include <stdio.h>#include <string.h>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <math.h>#include <algorithm>using namespace std;#define ls 2*i#define rs 2*i+1#define up(i,x,y) for(i=x;i<=y;i++)#define down(i,x,y) for(i=x;i>=y;i--)#define mem(a,x) memset(a,x,sizeof(a))#define w(a) while(a)#define LL long longconst double pi = acos(-1.0);#define Len 200005#define mod 1000000007const int INF = 0x3f3f3f3f;int vis[105][105];int father[105];int d[105],s[105];LL ans;int find(int x){    if(father[x]==x) return x;    return father[x] = find(father[x]);}int main(){    int n,m,i,j,k,x;    w(~scanf("%d%d",&n,&m))    {        mem(vis,0);        up(i,0,n)        {            father[i] = i;            s[i] = 1;//这棵树的节点的情况            d[i] = 0;//d[i]是i的周围必须安排几个人        }        int sum = n,flag = 0;        up(i,1,m)        {            scanf("%d",&x);            if(vis[i][x] || vis[x][i]) continue;            vis[i][x] = vis[x][i] = 1;            d[i]++;            d[x]++;            if(find(x)!=find(i))            {                s[find(i)]+=s[find(x)];                father[find(x)] = father[find(i)];                sum--;//减少了一棵树            }            else                flag = 1;//成环        }        if(n==2)//只有两个人必然只有一种坐法        {            printf("1\n",ans);            continue;        }        up(i,1,n)        {            if(d[i]>2)//i的旁边要安排的人数大于2，必然无法满足                break;        }        if(i<=n || (sum>1&&flag))//成环并且除了这棵子树还有其他子树，必然无法满足        {            printf("0\n");            continue;        }        ans = 1;        up(i,1,sum-1)//对于这些树全排列的方式有几种，但是1的位置是固定的        ans = (ans*i)%mod;        up(i,1,n)        {            if(father[i]==i && s[i]>1)//看根节点，如果这棵树不只一个节点，那么就有两种情况                ans = (ans*2)%mod;        }        printf("%I64d\n",ans);    }    return 0;}