[leetcode] 1192 critical-connections-in-a-network
作者:互联网
https://leetcode.com/contest/weekly-contest-154/problems/critical-connections-in-a-network/
class Solution { public: vector<vector<int>> res; int index = 0; // 访问序号 vector<int> dfn; // 当前访问的序号 vector<int> low; // 当前节点&其子树 最早访问的序号 vector<vector<int>> graph; void tarjan(int cur, int last) { low[cur] = dfn[cur] = index++; // 分配一个序号 for (const auto& next : graph[cur]) { if(next == last) continue; // 避免重复访问 if (dfn[next] == -1) { tarjan(next, cur); // 没有访问过,继续搜索 low[cur] = min(low[cur], low[next]); // 更新最早序号 if(low[next] > dfn[cur]) // 新的节点的low已经大于当前节点的序号,说明已经不在同一个强联通分量里了 { res.push_back({cur, next}); // 加入到结果中 } } else { low[cur] = min(low[cur], dfn[next]);// 访问过了,直接更新 //low[cur] = min(low[cur], low[next]); -> I think this is right ! } } } vector<vector<int>> criticalConnections(int n, vector<vector<int>>& connections) { dfn.resize(n, -1); low.resize(n, -1); graph.resize(n); for (int i = 0; i < connections.size(); i++) { vector<int>& c = connections[i]; int v1 = c[0]; int v2 = c[1]; graph[v1].push_back(v2); graph[v2].push_back(v1); } tarjan(0, -1); // 处在同一个强联通分量中节点, 最终会得到相同的low值 // for(int i = 0;i < n;i++) cout << low[i] << " " << dfn[i] << endl; return res; } };
标签:vector,cur,int,1192,next,connections,critical,low,dfn 来源: https://www.cnblogs.com/fish1996/p/11522072.html