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 <bits/stdc++.h>
#ifdef __DEBUG__
    #define DBG(X) cout << #X << " = " << (X) << endl;
    #define SAY(X) cout << (X) << endl;
#else
    #define DBG(X)
    #define SAY(X)
#endif

using namespace std;

using ll = long long int;
using ull = unsigned long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;

int dx[]={1, 0, -1, 0, 1,-1, 1,-1};
int dy[]={0, 1, 0, -1, 1, 1,-1,-1};
const int INT_INF = (int)(2e9);
const ll  LL_INF = (ll)(2e18);

inline void fast_io() { ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); };
template<typename T, typename S> inline ostream& operator<<(ostream& os, const pair<T, S> p) { cout << "[" << p.first << ";" << p.second << "]"; return os; }
template<typename T, typename S> inline ostream& operator<<(ostream& os, const map<T, S> p) { for (auto el : p) cout << "[" << el.first << ";" << el.second << "]"; return os; }
template<typename T> inline ostream& operator<<(ostream& os, const vector<T>& v) { for (auto el : v) cout << el << " "; return os; }
template<typename T> inline ostream& operator<<(ostream& os, const deque<T>& v) { for (auto el : v) cout << el << " "; return os; }
template<typename T> inline ostream& operator<<(ostream& os, const set<T>& v) { for (auto el : v) cout << el << " "; return os; }
template<typename T> inline vector<T> fetch_vec(int sz) { vector<T> ret(sz); for (auto& elem : ret) cin >> elem; return ret; }

int N,M;
vector<vector<int>> G,revG;
vector<pii> Edge;
void input(){
    fast_io();
    cin >> N >> M;
    G.resize(N);
    revG.resize(N);
    for (int i = 0; i < M; i++)
    {
        int s,t; cin >> s >> t;
        s--,t--;
        Edge.emplace_back(s,t);
        G[s].push_back(t);
        revG[t].push_back(s);
    }
    Edge.emplace_back(N,N);
}

vector<double> memo;
double dfs(int s,int i,const vector<vector<int>> &g){
    if(memo[s]) return memo[s];
    if(s==N-1) return memo[s] = 0;
    if(g[s].empty()) return memo[s] = N+1;
    if(s==Edge[i].first && g[s].size()==1) return N+1;
    double ret = 1;
    for(int v:g[s]){
        if(s==Edge[i].first && v==Edge[i].second) continue;
        if(s!=Edge[i].first) ret += dfs(v,i,g)/g[s].size();
        else ret += dfs(v,i,g)/(g[s].size()-1);
    }
    return memo[s] = ret;
}

// 繋がってなかったらdfsの値が∞になるようにしてるから意味ない。
bool isok(int i,vector<vector<int>> g){
    queue<int> q;
    q.push(0);
    vector<bool> seen(N,false);
    seen[0] = true;
    while (!q.empty())
    {
        int u = q.front();
        q.pop();
        for(int v:g[u]){
            if(u==Edge[i].first && v==Edge[i].second) continue;
            if(seen[v]) continue;
            if(v==N-1) return true;
            seen[v] = true;
            q.push(v);
        }
    }
    return seen[N-1];
}

int TLEsolve(){
    double ans = N+1;
    for (int i = 0; i <= M; i++)
    {
        // if(isok(i,G)) 
        memo.assign(N,0);
        ans = min(ans,dfs(0,i,G));
    }
    cout << fixed << setprecision(10) << ans << endl;    
    return 0;
}   

double eps = 1e-9;
int TLEsolve2(){
    vector<vector<double>> dp(N+1,vector<double>(N+1,0));
    dp[0][0] = 1;
    for (int i = 0; i < N; i++)
    {
        for (int X = 0; X < N; X++)
        {
            if(dp[i][X]==0) continue;
            for(int u:G[i]){
                dp[u][X+1] += dp[i][X]/G[i].size();
            }
        }        
    }
    double ans = 0;
    DBG(dp[N-1])
    double Pmod = 0;
    for (int x = 1; x < N; x++)
    {
        Pmod += dp[N-1][x];
    }

    for (int x = 1; x < N; x++)
    {
        dp[N-1][x]/= Pmod;
        ans += dp[N-1][x]*x;
    }

    for (int k = 0; k < M; k++)
    {
        int s = Edge[k].first;
        int t = Edge[k].second;
        if(G[s].size()==1) continue;
        vector<vector<double>> dp_diff(N+1,vector<double>(N+1,0));

        for (int X = 0; X < N; X++)
        {
            if(dp[s][X]==0) continue;
            for(int u:G[s]){
                if(u==t) dp_diff[u][X+1] -= dp[s][X]/G[s].size();
                else dp_diff[u][X+1] += dp[s][X]/(G[s].size()-1)-dp[s][X]/G[s].size();
            }
        }
        
        for (int i = s+1; i < N; i++)
        {
            for (int X = 0; X < N; X++)
            {
                if(dp_diff[i][X]==0) continue;
                for(int u:G[i]){
                    dp_diff[u][X+1] += dp_diff[i][X]/G[i].size();
                }
            }        
        }

        double ans_tmp = 0;
        DBG(dp_diff[N-1])

        for (int x = 1; x < N; x++)
        {
            ans_tmp += (dp[N-1][x]+dp_diff[N-1][x])*x;
        }
        if(abs(ans_tmp)<=eps) continue;
        DBG(ans_tmp)
        ans = min(ans,ans_tmp);
    }
    cout << fixed << setprecision(10) << ans << endl;
    return 0;
}

double solveG(int y, vector<vector<int>> g){
    vector<double> dp(N+1,0);
    vector<double> pr(N+1,0);
    dp[0] = 0;
    pr[0] = 1;
    for (int u = 0; u < N; u++)
    {
        if(pr[u]==0) continue;
        for(int v:g[u]){
            if(u==Edge[y].first && v==Edge[y].second) continue;
            if(u==Edge[y].first) pr[v] += pr[u]/(g[u].size()-1);
            else pr[v] += pr[u]/g[u].size();
        }
    }


    for (int u = 0; u < N; u++)
    {
        for(auto v:g[u]){
            if(u==Edge[y].first && v==Edge[y].second) continue;
            if(u==Edge[y].first) dp[v] += dp[u]/(g[u].size()-1)+pr[u]/(g[u].size()-1);
            else dp[v] += dp[u]/g[u].size()+pr[u]/g[u].size();
        }
    }
    return dp[N-1];
}

int solve(){
    double ans = N+1;
    for (int i = 0; i <= M; i++)
    {
        if(G[Edge[i].first].size()==1) continue;
        double tmp = solveG(i,G);
        ans = min(ans,tmp);
    }
    cout << fixed << setprecision(10) << ans << endl;
    return 0;
}

int main()
{
    input();
    solve();
    return 0;
}