有向无环图随机游走(Random Walk on DAG) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <iostream>
#include <queue>
#include <vector>
using namespace std;

const int MOD = 998244353;

long long qpow(long long a, long long b) {
    long long res = 1;
    while (b > 0) {
        if (b & 1) res = res * a % MOD;
        a = a * a % MOD;
        b >>= 1;
    }
    return res;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, m;
    if (!(cin >> n >> m)) return 0;
    vector<vector<int>> adj(n + 1);
    vector<int> indeg(n + 1, 0);
    for (int i = 0; i < m; ++i) {
        int u, v;
        cin >> u >> v;
        adj[u].push_back(v);
        indeg[v]++;
    }
    queue<int> q;
    vector<int> ord;
    ord.reserve(n);
    for (int i = 1; i <= n; ++i) {
        if (indeg[i] == 0) q.push(i);
    }
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        ord.push_back(u);
        for (int v : adj[u]) {
            if (--indeg[v] == 0) q.push(v);
        }
    }
    vector<long long> cnt(n + 1, 1), sum(n + 1, 0);
    for (int i = n - 1; i >= 0; --i) {
        int u = ord[i];
        for (int v : adj[u]) {
            cnt[u] = (cnt[u] + cnt[v]) % MOD;
            sum[u] = (sum[u] + sum[v] + cnt[v]) % MOD;
        }
    }
    long long tot_cnt = 0, tot_sum = 0;
    for (int i = 1; i <= n; ++i) {
        tot_cnt = (tot_cnt + cnt[i]) % MOD;
        tot_sum = (tot_sum + sum[i]) % MOD;
    }
    cout << tot_sum * qpow(tot_cnt, MOD - 2) % MOD << "\n";
    return 0;
}