摘要: //割边 int dfn[MAXN], low[MAXN], idx; bool bridge[MAXN]; void tarjan(int now, int f) { dfn[now] = low[now] = ++idx; for (int i = head[now]; i; i = nxt[i 阅读全文
posted @ 2020-07-14 15:15 行zzz 阅读(142) 评论(0) 推荐(0)