Problem:
Score : 600 points

Problem Statement

There is a cave consisting of N rooms and M one-directional passages. The rooms are numbered 1 through N.

Takahashi is now in Room 1, and Room N has the exit. The i-th passage connects Room s_i and Room t_i (s_i < t_i) and can only be traversed in the direction from Room s_i to Room t_i. It is known that, for each room except Room N, there is at least one passage going from that room.

Takahashi will escape from the cave. Each time he reaches a room (assume that he has reached Room 1 at the beginning), he will choose a passage uniformly at random from the ones going from that room and take that passage.

Aoki, a friend of Takahashi's, can block one of the passages (or do nothing) before Takahashi leaves Room 1. However, it is not allowed to block a passage so that Takahashi is potentially unable to reach Room N.

Let E be the expected number of passages Takahashi takes before he reaches Room N. Find the value of E when Aoki makes a choice that minimizes E.

Constraints

2 \leq N \leq 600

N-1 \leq M \leq \frac{N(N-1)}{2}

s_i < t_i

If i != j, (s_i, t_i) \neq (s_j, t_j). (Added 21:23 JST)

For every v = 1, 2, ..., N-1, there exists i such that v = s_i.

Input

Input is given from Standard Input in the following format:

N M
s_1 t_1
:
s_M t_M

Output

Print the value of E when Aoki makes a choice that minimizes E.
Your output will be judged as correct when the absolute or relative error from the judge's output is at most 10^{-6}.

Sample Input 1

4 6
1 4
2 3
1 3
1 2
3 4
2 4

Sample Output 1

1.5000000000

If Aoki blocks the passage from Room 1 to Room 2, Takahashi will go along the path 1 → 3 → 4 with probability \frac{1}{2} and 1 → 4 with probability \frac{1}{2}. E = 1.5 here, and this is the minimum possible value of E.

Sample Input 2

3 2
1 2
2 3

Sample Output 2

2.0000000000

Blocking any one passage makes Takahashi unable to reach Room N, so Aoki cannot block a passage.

Sample Input 3

10 33
3 7
5 10
8 9
1 10
4 6
2 5
1 7
6 10
1 4
1 3
8 10
1 5
2 6
6 9
5 6
5 8
3 6
4 8
2 7
2 9
6 7
1 2
5 9
6 8
9 10
3 9
7 8
4 5
2 10
5 7
3 5
4 7
4 9

Sample Output 3

3.0133333333

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <iomanip>
#include <set>
using namespace std;

set<int> Adj[606];

int N, M;

int s[360005], t[360005];

int C[605];

vector<int > V;

int used[605];


long long int val[605];

int num[605];

long double p[605];



long double E[605];

vector<int> c_Adj[605];

long double a[605], b[605], x[605];

int main(void) {

	cin >> N >> M;

	for (int i = 1; i <= M; i++) {
		cin >> s[i] >> t[i];
		Adj[s[i]].insert(t[i]);
		num[t[i]]++;
		c_Adj[t[i]].push_back(s[i]);
	}

	long double ans = 999999999999999.9;

	//==================================
	vector<int> W;

	for (int k = 1; k <= N; k++) {
		int q;
		for (int i = 1; i <= N; i++) {
			if (num[i] == 0) { q = i; break; }
		}

		num[q] = -1;
		for (auto y : Adj[q]) { num[y]--; }
		W.push_back(q);
	}

	//for (auto y : W) { cout << " " << y; }cout << endl;

	s[0] = 0; t[0] = 0;



	//============================

	for (int i = 1; i <= N; i++) { E[i] = -1.0L; }

	E[N] = 0.0;

	for (int i = W.size() - 1; i >= 0; i--) {

		int Y = W[i];
		int cnt = 0;
		long double total = 0.0L;
		for (auto z : Adj[Y]) {
			if (E[z] < 0.0L) {
			}
			else {
				cnt++;
				total += E[z];
			}
		}

		if (cnt > 0) { E[Y] = total / cnt + 1.0L; }
	}

	ans = min(ans, E[1]);


	long double sum = E[1];

	for (int i = 1; i <= N; i++) { x[i] = E[i]; }

	//==========================
	//  b
	//=======================
	for (int k = 1; k <= N; k++) {

		for (int i = 1; i <= N; i++) { E[i] = -1.0L; }

		E[N] = 0.0;

		for (int i = W.size() - 1; i >= 0; i--) {

			int Y = W[i];
			//if (Y == k) { E[k] = 0.0; continue; }
			int cnt = 0;
			long double total = 0.0L;
			for (auto z : Adj[Y]) {
					cnt++;
					total += E[z];
			}

			if (cnt > 0) { E[Y] = total / cnt + 1.0L; }

			if (Y == k) { E[k] += 1.0L; }
		}

		//b[k] = E[1];
		//a[k] = (sum - b[k]) / x[k];
		a[k] = E[1] - sum;
		b[k] = sum - a[k] * x[k];

	}// k-fin

	/*for (int i = 1; i <= N; i++) {
		cout << "a[" << i << "]= " << a[i] << "  x[" << i << "]= " << x[i] << " b[" << i << "]=  " << b[i] << endl;
	}*/

	//===========================

	for (int k = 1; k <= M; k++) {
		if (Adj[s[k]].size() <= 1) { continue; }
		//cout << s[k] << "  " << t[k] << endl;

		int cnt = 0;
		long double tmp = 0.0;
		for (auto u : Adj[s[k]]) {
			if (u == t[k]) { continue; }
			cnt++;
			tmp += x[u];
		}
		long double value = a[s[k]]*(tmp/cnt+1.0L) + b[s[k]];
		//cout << "value= " << value << endl;
		ans = min(ans, value);
	}


/*	for (int k = 0; k <= M; k++) {

		if (Adj[s[k]].size() <= 1) { continue; }
	
		Adj[s[k]].erase(t[k]);

		for (int i = 1; i <= N; i++) { C[i] = 0; val[i] = 0; E[i] = 0; }


		for (int i = 1; i <= N; i++) { E[i] = -1.0L; }

		E[N] = 0.0;

		for (int i = W.size() - 1; i >= 0; i--){

			int Y = W[i];
			int cnt = 0;
			long double total = 0.0L;
			for (auto z : Adj[Y]) {
				if (E[z] < 0.0L) {
				}
				else {
					cnt++;
					total += E[z];
				}
			}

			if (cnt > 0) { E[Y] = total / cnt + 1.0L; }
      }


		ans = min(ans, E[1]);

		Adj[s[k]].insert(t[k]);
	}
	*/
	//=============================================



	cout << setprecision(12)<<fixed<<ans << endl;
	return 0;
}