我做的是这道题 最小路径覆盖问题 （网络流），代码中使用的是 Dinic 算法，程序在最后一行异常退出，详见 提交记录。

#include <iostream>
#include <queue>
#include <cstring>
#include <climits>
#include <map>

using namespace std;
const int Max = 2000;

int n, m, s, t, ans, flow, dis[Max], to[Max];
map<int, int> g[Max];
bool d[Max];

bool bfs() {
    memset(dis, 0, sizeof dis);
    queue<int> q;
    q.push(s);
    dis[s] = 1;

    while (!q.empty()) {
        auto x = q.front();
        q.pop();
        if (x == t)
            return true;

        for (auto y: g[x])
            if (y.second && !dis[y.first]) {
                dis[y.first] = dis[x] + 1;
                q.push(y.first);
            }
    }
    return false;
}

int dinic(int x, int _flow) {
    if (x == t)
        return _flow;
    int rest = _flow;

    for (auto y: g[x])
        if (rest && y.second && dis[y.first] == dis[x] + 1) {
            int f = dinic(y.first, min(rest, y.second));
            if (!f)
                dis[y.first] = -1;
            else {
                if (x != s)
                    d[y.first - n] = true;
                to[x] = y.first;
                rest -= f;
                g[x][y.first] -= f;
                g[y.first][x] += f;
            }
        }
    return _flow - rest;
}

int main() {
    cin >> n >> m;
    s = 0;
    t = n + n + 1;

    for (int i = 1; i <= n; i++) {
        g[s][i] = 1;
        g[i][s] = 0;
        g[i + n][t] = 1;
        g[t][i + n] = 0;
    }

    for (int i = 1, x, y; i <= m; i++) {
        cin >> x >> y;
        g[x][y + n] = 1;
        g[y + n][x] = 0;
    }

    while (bfs()) {
        while ((flow = dinic(s, INT_MAX)))
            ans += flow;
    }

    for (int i = 1; i <= n; i++)
        if (!d[i]) {
            int x = i;
            cout << x;
            while (to[x] && to[x] != t) {
                x = to[x] - n;
                cout << ' ' << x;
            }
            cout << endl;
        }

    cout << n - ans << endl;
    return 0;
}