五项一起对得上吗

运行下面这段程序：

本节的图：七个点，边 0-1、0-2、1-3、2-4、3-4、5-6；邻接表 G 里每个点的邻居按编号从小到大排。

六门课的先修关系：0→2、1→2、2→3、2→4、3→5、4→5（邻接表 COURSE）。

#include <algorithm>
#include <iostream>
#include <queue>
#include <string>
#include <utility>
#include <vector>
using namespace std;

vector<vector<int>> G = {{1, 2}, {0, 3}, {0, 4}, {1, 4}, {2, 3}, {6}, {5}};

void go(const vector<vector<int>>& g, int u, vector<bool>& seen, vector<int>& out) {
    seen[u] = true;
    out.push_back(u);
    for (int v : g[u]) {
        if (!seen[v]) go(g, v, seen, out);
    }
}

vector<int> dfs(const vector<vector<int>>& g, int s) {
    vector<bool> seen(g.size(), false);
    vector<int> out;
    go(g, s, seen, out);
    return out;
}

vector<int> bfs(const vector<vector<int>>& g, int s) {
    vector<bool> seen(g.size(), false);
    queue<int> q;
    vector<int> out;
    seen[s] = true;
    q.push(s);
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        out.push_back(u);
        for (int v : g[u]) {
            if (!seen[v]) {
                seen[v] = true;
                q.push(v);
            }
        }
    }
    return out;
}

vector<vector<int>> blocks(const vector<vector<int>>& g) {
    vector<bool> seen(g.size(), false);
    vector<vector<int>> out;
    for (int s = 0; s < (int)g.size(); s++) {
        if (seen[s]) continue;
        vector<int> blk;
        vector<int> st = {s};
        seen[s] = true;
        while (!st.empty()) {
            int u = st.back();
            st.pop_back();
            blk.push_back(u);
            for (int v : g[u]) {
                if (!seen[v]) {
                    seen[v] = true;
                    st.push_back(v);
                }
            }
        }
        sort(blk.begin(), blk.end());
        out.push_back(blk);
    }
    return out;
}

vector<int> dist(const vector<vector<int>>& g, int s) {
    vector<int> d(g.size(), -1);
    queue<int> q;
    d[s] = 0;
    q.push(s);
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        for (int v : g[u]) {
            if (d[v] == -1) {
                d[v] = d[u] + 1;
                q.push(v);
            }
        }
    }
    return d;
}

vector<vector<int>> COURSE = {{2}, {2}, {3, 4}, {5}, {5}, {}};

vector<int> kahn(const vector<vector<int>>& g) {
    vector<int> d(g.size(), 0);
    for (size_t u = 0; u < g.size(); u++) {
        for (int v : g[u]) d[v]++;
    }
    queue<int> q;
    for (int u = 0; u < (int)g.size(); u++) {
        if (d[u] == 0) q.push(u);
    }
    vector<int> out;
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        out.push_back(u);
        for (int v : g[u]) {
            if (--d[v] == 0) q.push(v);
        }
    }
    return out;
}

int main() {
    bool ok = dfs(G, 0) == vector<int>{0, 1, 3, 4, 2}
        && bfs(G, 0) == vector<int>{0, 1, 2, 3, 4}
        && blocks(G).size() == 2
        && dist(G, 0)[4] == 2
        && kahn(COURSE).size() == 6;
    cout << boolalpha << ok << endl;
}

（本题用 g++ -std=c++17 -O0 编译。）