ZOJ-3319

Islands

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Time Limit: 1 Second      Memory Limit: 32768 KB

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There are N islands and some directed paths between some of them. You are the transportation minister of these islands, and you are going to build some more directed paths so that every island belongs to exactly one cycle. A cycle is two or more islands I₁, I₂, I₃, ... I_k, such that there are paths: I₁ -> I₂, I₂ -> I₃, ... and I_k -> I₁. Besides the cycles, there should not be any extra edges. Of course, you cannot build a path from an island to itself. You want to calculate how many different ways you can build paths to satisfy the restriction.

Input

There are multiple cases (no more than 100). For each case, the first line is an integer N (1 <= N <= 100), giving the number of islands. N = 0 indicates the end of input. Then N lines follow, each with N characters, giving the paths between islands. The j-th character of the i-th line is either 'Y' or 'N'. 'Y' means there is a path from the i-th island to the j-th island, while 'N' means there is no path from the i-th island to the j-th island. The i-th character of the i-th line is always 'N'.

Output

For each case, you should output how many different ways you can build paths to satisfy the restriction. The answer may be very large, so just output the answer MOD 10,000,007.

Sample Input

2
NN
NN
2
NY
YN
3
NNN
NNN
NNN
3
NYY
NNN
NNN
0

Sample Output

1
1
2
0

---

Author: HANG, Hang
Source: The 10th Zhejiang University Programming Contest

/**
          题意：给出一个有向图，然后让所有点在一个环内；
          做法：组合数，错排
                    错排，当加一个点在链中时，可以将链看成是一个点，然后进行排列的方法有多少种
                    组合数，链可以是一条链加点成环，也可以时多条链加点成环。
**/
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <queue>
#include <set>
#include <vector>
#define mod 10000007
#define maxn 110
using namespace std;
char ch[maxn][maxn];
int indegree[maxn];
int outdegree[maxn];
int cuo[maxn];
long long  C[maxn][maxn];
int n;
void  init()
{
    C[0][0] = 1;
    for(int i = 1 ; i <= 100 ; i++)
    {
        C[i][0] = C[i][i] = 1;
        for(int j = 1; j < i; j++)
        {
            C[i][j] = (C[i-1][j] + C[i-1][j-1] )%mod;
        }
    }
    cuo[0] = 1;
    cuo[1] = 0;
    for(int i=2; i<=100; i++)  ///错排
    {
        cuo[i] = ((i-1) *(cuo[i-1] + cuo[i-2]))%mod;
    }
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
    init();
    while(~scanf("%d",&n))
    {
        if(n == 0) break;
        memset(indegree,0,sizeof(indegree));
        memset(outdegree,0,sizeof(outdegree));
        for(int i=0; i<n; i++)
        {
            scanf("%s",ch[i]);
            for(int j=0; j<n; j++)
            {
                if(ch[i][j] == 'Y')
                {
                    indegree[j] ++;
                    outdegree[i]++;
                }
            }
        }
        bool prime = true;
        for(int i=0; i<n; i++)
        {
            if(indegree[i] > 1 || outdegree[i] >1)
            {
                prime = false;
                break;
            }
        }
        if(!prime)
        {
            printf("0
");      ///原图存在环
        }
        else
        {
            int In= 0,Out = 0,tot = 0;
            for(int i=0; i<n; i++)
            {
                if(indegree[i] == 0)
                {
                    In++;
                    if(outdegree[i] == 0)   
                    {
                        Out++;              ///孤立的点的个数
                    }
                }
            }
            tot = In - Out;    ///弧的个数
            long long ans = 0;
            for (int i = 0; i <= tot; i++)
            {
                ans = (ans + C[tot][i]*cuo[i+Out])%mod;
            }
            printf("%lld
", ans);
        }
    }
    return 0;
}