BCCs and 2CCs
Authors: Benjamin Qi, Mihnea Brebenel
Resources | ||||
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CF | ||||
CP2 |
2-Edge-Connected Components
Status | Source | Problem Name | Difficulty | Tags | |
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YS | Easy | Show Tags2CC |
C++
#include <bits/stdc++.h>using namespace std;const int MAX_N = 3e5;int timer; // Time of entry in nodeint scc; // Number of strongly connected componentsint id[MAX_N];int low[MAX_N]; // Lowest ID in node's subtree in DFS treevector<int> neighbors[MAX_N];
This section is not complete.
(implementation)
With DSU
Status | Source | Problem Name | Difficulty | Tags | |
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Platinum | Normal | Show TagsMerging |
The analysis for the above problem mentions an solution. Although this is not a two-connected component problem, we can in fact use DSU to generate two-connected components.
This section is not complete.
(explanation?)
The DSU operations take rather than because the DSU does not use union by size, but it's easy to change this.
struct TwoEdgeCC {struct {vi e;void init(int n) { e = vi(n, -1); }int get(int x) { return e[x] < 0 ? x : e[x] = get(e[x]); }bool unite(int x, int y) { // set par[y] = xx = get(x), y = get(y);if (x == y) return 0;e[x] += e[y];e[y] = x;
Problems
Focus Problem – try your best to solve this problem before continuing!
- SRM 787 1000
Biconnected Components
Focus Problem – try your best to solve this problem before continuing!
An important observation is that if you can't go from node to node without passing through node , node is a critical node (articulation point). Node can split and into 2 different components if removed. This makes us think about biconnected components.
Now we're left with two cases. If node isn't critical, a path from to can avoid the node. Otherwise, if node is a critical one we have to check if is on path from to . Here is a little tricky, on a simple graph, it's not so easy to check, on the other hand, checking this on a tree can be much easier. In order to do this, we transform the graph into a block-cut tree.
In a block-cut tree, every articulation and biconnected component represents a node. Now that we have turned our graph into a tree how do we check if node the path from to passes through ? To do this we use LCA. You can find more about this here.
C++
#include <bits/stdc++.h>using namespace std;/** @return the block-cut tree of a graph */vector<vector<int>> biconnected_components(vector<vector<int>> &g,vector<bool> &is_cutpoint,vector<int> &id) {int n = (int)g.size();vector<vector<int>> comps;
Articulation Points
Focus Problem – try your best to solve this problem before continuing!
Resources | ||||
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CP2 | maybe not completely correct |
C++
#include <bits/stdc++.h>using namespace std;const int NMAX = 2e4;int timer;vector<int> low, id;vector<bool> visited, ap;vector<vector<int>> g(NMAX);
Problems
Status | Source | Problem Name | Difficulty | Tags | |
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CSES | Easy | Show TagsBCC | |||
POI | Normal | Show TagsBCC | |||
APIO | Normal | Show TagsBCC | |||
POI | Normal | Show TagsBCC | |||
TLE | Hard | Show TagsBCC | |||
CEOI | Hard | Show TagsBCC | |||
Platinum | Very Hard | Show TagsBCC |
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