题解 | #绿豆蛙的归宿#

首先拓扑排序

然后定义状态表示表示从当前点到达的期望, 根据无后效性原理, 从后向前即可

#include <bits/stdc++.h>

#define x first
#define y second
#define all(x) x.begin(), x.end()
#define vec1(T, name, n, val) vector<T> name(n, val)
#define vec2(T, name, n, m, val) vector<vector<T>> name(n, vector<T>(m, val))
#define vec3(T, name, n, m, k, val) vector<vector<vector<T>>> name(n, vector<vector<T>>(m, vector<T>(k, val)))
#define vec4(T, name, n, m, k, p, val) vector<vector<vector<vector<T>>>> name((n), vector<vector<vector<T>>>((m), vector<vector<T>>((k), vector<T>((p), (val)))))

using namespace std;
using i128 = __int128;
using u128 = unsigned __int128;
using LL = long long;
using LD = long double;
using ULL = unsigned long long;
using PII = pair<int, int>;
using PLL = pair<LL, LL>;
using PLD = pair<LD, LD>;

const int N = 1e5 + 10, MOD = 1e9 + 7;
const int INF = 1e9;
const LL LL_INF = 2e18;
const LD EPS = 1e-8;
const int dx4[] = {-1, 0, 1, 0}, dy4[] = {0, 1, 0, -1};
const int dx8[] = {-1, -1, -1, 0, 0, 1, 1, 1}, dy8[] = {-1, 0, 1, -1, 1, -1, 0, 1};

istream& operator>>(istream& is, i128& val) {
    string str;
    is >> str;
    val = 0;
    bool flag = false;
    if (str[0] == '-') flag = true, str = str.substr(1);
    for (char& c : str) val = val * 10 + c - '0';
    if (flag) val = -val;
    return is;
}

ostream& operator<<(ostream& os, i128 val) {
    if (val < 0) os << "-", val = -val;
    if (val > 9) os << val / 10;
    os << static_cast<char>(val % 10 + '0');
    return os;
}

bool cmp(LD a, LD b) {
    if (fabs(a - b) < EPS) return 1;
    return 0;
}

LL qpow(LL a, LL b) {
    LL ans = 1;
    a %= MOD;
    while (b) {
        if (b & 1) ans = ans * a % MOD;
        a = a * a % MOD;
        b >>= 1;
    }
    return ans;
}

void solve() {
    int n, m;
    cin >> n >> m;
    vector<vector<PII>> g(n + 1);
    vector<int> ideg(n + 1), odeg(n + 1);

    for (int i = 0; i < m; ++i) {
        int a, b, c;
        cin >> a >> b >> c;
        g[a].push_back({b, c});
        ideg[b]++;
        odeg[a]++;
    }

    queue<int> q;
    vector<int> b;
    b.reserve(n + 1);
    for (int i = 1; i <= n; ++i) {
        if (ideg[i] == 0) {
            q.push(i);
            b.push_back(i);
        }
    }

    while (q.size()) {
        int u = q.front();
        q.pop();
        for (auto& [v, w] : g[u]) {
            ideg[v]--;
            if (ideg[v] == 0) {
                q.push(v);
                b.push_back(v);
            }
        }
    }

    vector<LD> f(n + 1);
    reverse(all(b));
    for (int u : b) {
        for (auto &[v, w] : g[u]) {
            f[u] += 1.0l * (f[v] + w) / odeg[u];
        }
    }

    cout << fixed << setprecision(2) << f[1] << '\n';

/**/ #ifdef LOCAL
    cout << flush;
/**/ #endif
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int T = 1;
    while (T--) solve();
    cout << fixed << setprecision(15);

    return 0;
}