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;
}