【题解】LuoGu1726：上白泽慧音_上白泽慧音强不强-CSDN博客

本文链接：https://blog.csdn.net/ModestCoder_/article/details/98648562

本文详细介绍了如何使用Tarjan算法寻找图中强连通分量的最大值，并选取字典序最小的顶点作为输出。通过具体代码实现，展示了从输入数据到输出结果的完整过程。

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原题传送门
就是求强连通分量的最大值，并且里面的点字典序最小
那先跑一遍tarjan，然后按编号从小到大找到第一个所属强连通分量满足点数最大
那么这个编号所属的强连通分量就是我们要输出的强连通分量
Code：

#include <bits/stdc++.h>
#define maxn 50010
using namespace std;
struct Edge{
	int to, next;
}edge[maxn << 1];
int num, head[maxn], Index, dfn[maxn], low[maxn], vis[maxn], top, sta[maxn], tot, color[maxn];
int cnt[maxn], n, m;

inline int read(){
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

void addedge(int x, int y){ edge[++num] = (Edge) { y, head[x] }; head[x] = num; }

void tarjan(int u){
	dfn[u] = low[u] = ++Index;
	vis[u] = 1, sta[++top] = u;
	for (int i = head[u]; i; i = edge[i].next){
		int v = edge[i].to;
		if (!dfn[v]) tarjan(v), low[u] = min(low[u], low[v]); else
		if (vis[v]) low[u] = min(low[u], dfn[v]);
	}
	if (dfn[u] == low[u]){
		++tot;
		while (sta[top + 1] != u) vis[sta[top]] = 0, color[sta[top--]] = tot, ++cnt[tot];
	}
}

int main(){
	n = read(), m = read();
	for (int i = 1; i <= m; ++i){
		int x = read(), y = read(), opt = read();
		addedge(x, y);
		if (opt == 2) addedge(y, x);
	}
	for (int i = 1; i <= n; ++i) if (!dfn[i]) tarjan(i);
	int Max = 0;
	for (int i = 1; i <= tot; ++i) Max = max(Max, cnt[i]);
	printf("%d\n", Max);
	int node = 0;
	for (int i = 1; i <= n; ++i)
		if (!node && cnt[color[i]] == Max) node = color[i], printf("%d ", i); else
		if (node == color[i]) printf("%d ", i);
	return 0;
}