强连通点对计数(Come Back in One Piece) 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 <vector>
#include <algorithm>
using namespace std;

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);
    for (int i = 0; i < m; ++i) {
        int u, v;
        cin >> u >> v;
        adj[u].push_back(v);
    }
    vector<int> dfn(n + 1, 0), low(n + 1, 0), in_stk(n + 1, 0);
    vector<int> stk, call_u, call_idx;
    int timer = 0;
    long long ans = 0;
    for (int i = 1; i <= n; ++i) {
        if (!dfn[i]) {
            call_u.push_back(i);
            call_idx.push_back(0);
            dfn[i] = low[i] = ++timer;
            stk.push_back(i);
            in_stk[i] = 1;
            while (!call_u.empty()) {
                int u = call_u.back();
                int& idx = call_idx.back();
                if (idx < (int)adj[u].size()) {
                    int v = adj[u][idx++];
                    if (!dfn[v]) {
                        dfn[v] = low[v] = ++timer;
                        stk.push_back(v);
                        in_stk[v] = 1;
                        call_u.push_back(v);
                        call_idx.push_back(0);
                    } else if (in_stk[v]) {
                        low[u] = min(low[u], dfn[v]);
                    }
                } else {
                    if (low[u] == dfn[u]) {
                        long long cnt = 0;
                        while (true) {
                            int x = stk.back();
                            stk.pop_back();
                            in_stk[x] = 0;
                            cnt++;
                            if (x == u) break;
                        }
                        ans += cnt * (cnt - 1) / 2;
                    }
                    call_u.pop_back();
                    call_idx.pop_back();
                    if (!call_u.empty()) {
                        low[call_u.back()] = min(low[call_u.back()], low[u]);
                    }
                }
            }
        }
    }
    cout << ans << "\n";
    return 0;
}