原題連結

裸的\(tarjan\)找強連通分量，記錄最大強連通分量即可，注意字典序。

#include<cstdio>
using namespace std;
const int N = 5010;
const int M = 1e5 + 10;
int fi[N], di[M], ne[M], dfn[N], low[N], sta[N], bl[N], ma_si, ma_id, ma_mi, SCC, tp, ti, l;
bool v[N];
inline int re()
{
    int x = 0;
    char c = getchar();
    bool p = 0;
    for (; c < '0' || c > '9'; c = getchar())
        p |= c == '-';
    for (; c >= '0' && c <= '9'; c = getchar())
        x = x * 10 + c - '0';
    return p ? -x : x;
}
inline void add(int x, int y)
{
    di[++l] = y;
    ne[l] = fi[x];
    fi[x] = l;
}
inline int minn(int x, int y){ return x < y ? x : y; }
void tarjan(int x)
{
    int i, y;
    dfn[x] = low[x] = ++ti;
    sta[++tp] = x;
    v[x] = 1;
    for (i = fi[x]; i; i = ne[i])
        if (!dfn[y = di[i]])
        {
            tarjan(y);
            low[x] = minn(low[x], low[y]);
        }
        else
            if (v[y])
                low[x] = minn(low[x], dfn[y]);
    if (!(low[x] ^ dfn[x]))
    {
        int mi = 1e9, s = 0;
        SCC++;
        do
        {
            y = sta[tp--];
            v[y] = 0;
            bl[y] = SCC;
            mi = minn(mi, y);
            s++;
        } while (x ^ y);
        if (ma_si < s)
        {
            ma_si = s;
            ma_id = SCC;
            ma_mi = mi;
        }
        else
            if (!(ma_si ^ s) && ma_mi > mi)
            {
                ma_id = SCC;
                ma_mi = mi;
            }
    }
}
int main()
{
    int i, n, m, x, y, z;
    n = re();
    m = re();
    for (i = 1; i <= m; i++)
    {
        x = re();
        y = re();
        z = re();
        add(x, y);
        if (z ^ 1)
            add(y, x);
    }
    for (i = 1; i <= n; i++)
        if (!dfn[i])
            tarjan(i);
    printf("%d\n", ma_si);
    for (i = 1; i <= n; i++)
        if (!(bl[i] ^ ma_id))
            printf("%d ", i);
    return 0;
}