洛谷1726 上白泽慧音

原题链接

裸的\(tarjan\)找强连通分量，记录最大强连通分量即可，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;
}