[Graph Theory] dual-connection Summary

Dual connection Summary

Such problems are divided into edge-Dual-connection and point-Dual-connection.

EDGE connection

After the edge is dual-connected, the component that does not have a dual-connected connection is the cut edge. Therefore, you can reduce the node into a tree and convert the problem into a tree. The cut edge is defined: after this edge is removed, the graph will not be connected.

Basically, each of these questions has a solution, which is to calculate the dual-connected component and then scale down the vertex into a tree for operations.
Or, you can directly request the edge to be cut and perform operations related to the edge.

Template:

# Include <cstdio> # include <cstring> # include <cstdlib> # include <algorithm> # include <vector> using namespace STD; const int n = 10005; const int M = 20005; int n, m, Val [N]; struct edge {int U, V, ID; bool iscut; edge () {} edge (int u, int V, int ID) {This-> U = u; this-> V = V; this-> id = ID; this-> iscut = false ;}} edge [M * 2], cut [m]; int en, first [N], next [m], cutn; void Init () {memset (first,-1, sizeof (first); e N = 0;} void add_edge (int u, int V, int ID) {edge [En] = edge (u, v, ID ); next [En] = first [u]; first [u] = EN ++;} int pre [N], dfn [N], bccno [N], BCCN, dfs_clock; void dfs_cut (int u, int FA) {pre [u] = dfn [u] = ++ dfs_clock; For (INT I = first [u]; I + 1; I = next [I]) {int v = edge [I]. v; If (edge [I]. id = FA) continue; If (! Pre [v]) {dfs_cut (v, edge [I]. ID); dfn [u] = min (dfn [u], dfn [v]); If (dfn [v]> pre [u]) {edge [I]. iscut = edge [I ^ 1]. iscut = true; // mark the cut edge cut [cutn ++] = edge [I]; // save cut edge} else dfn [u] = min (dfn [u], pre [v]);} void find_cut () {dfs_clock = cutn = 0; memset (PRE, 0, sizeof (pre); For (INT I = 0; I <n; I ++) if (! Pre [I]) dfs_cut (I,-1);} void dfs_bcc (INT U) {pre [u] = 1; bccno [u] = BCCN; for (INT I = first [u]; I + 1; I = next [I]) {If (edge [I]. iscut) continue; int v = edge [I]. v; If (pre [v]) continue; dfs_bcc (v) ;}} vector <int> BCC [N]; void find_bcc () {BCCN = 0; memset (PRE, 0, sizeof (pre); For (INT I = 0; I <n; I ++) {If (! Pre [I]) {dfs_bcc (I); BCCN ++ ;}} int main () {return 0 ;}

HDU 2242 road to postgraduate entrance-air-conditioning classrooms with dual-connection deflation points + DFS
HDU 2460 Network side dual-connection deflation point + tree link splitting
HDU 3849by recognizing these guys, we find social networks useful bridge
HDU 4005 the war side dual-connection deflation point + DFS
HDU 3394 railway dual-connection Block
3177 redundant paths constructs edge dual-connectivity and uses the degree to calculate
3352 Road Construction
1515 street directions can be changed to an undirected graph, and the bridge cannot be changed.
1438 one-way traffic hybrid graph change directed graph (the data in this question seems a bit problematic)

Point dual-connection

Point dual-connectivity, find out, connect to the cut point, note that a cut point may belong to different points-Dual-connectivity component, the cut point is defined as: After removing this point, graph will not be connected

Template:

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>#include <stack>using namespace std;const int N = 1005;const int M = 2000005;int n, m;struct Edge {int u, v;Edge() {}Edge(int u, int v) {this->u = u;this->v = v;}} edge[M];int first[N], next[M], en;void add_edge(int u, int v) {edge[en] = Edge(u, v);next[en] = first[u];first[u] = en++;}void init() {en = 0;memset(first, -1, sizeof(first));}int pre[N], dfn[N], bccno[N], dfs_clock, bccn;bool iscut[N];stack<Edge> S;vector<int> bcc[N];void dfs_bcc(int u, int fa) {dfn[u] = pre[u] = ++dfs_clock;int child = 0;for (int i = first[u]; i + 1; i = next[i]) {int v = edge[i].v;if (fa == v) continue;if (!pre[v]) {S.push(edge[i]);child++;dfs_bcc(v, u);dfn[u] = min(dfn[u], dfn[v]);if (dfn[v] >= pre[u]) {iscut[u] = true;bccn++;bcc[bccn].clear();while (1) {Edge x = S.top(); S.pop();if (bccno[x.u] != bccn) {bcc[bccn].push_back(x.u); bccno[x.u] = bccn;}if (bccno[x.v] != bccn) {bcc[bccn].push_back(x.v); bccno[x.v] = bccn;}if (x.u == u && x.v == v) break;}}} else if (pre[v] < pre[u] && v != fa) {S.push(edge[i]);dfn[u] = min(dfn[u], pre[v]);}}if (fa == 0 && child == 1) iscut[u] = 0;}void find_bcc() {memset(pre, 0, sizeof(pre));memset(bccno, 0, sizeof(bccno));memset(iscut, false, sizeof(iscut));dfs_clock = bccn = 0;for (int i = 1; i <= n; i++)if (!pre[i]) dfs_bcc(i, 0);}

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Ultraviolet A 1108 mining your own business point dual-Connected Component + count

[Graph Theory] dual-connection Summary