EBCC
- 例题
PBCC
- 涉及割点/PBCC时的解决思路
- 例题
SCC
- 过去分析
- 例题

EBCC

Some properties of the algorithm:
+ for a connected-graph GGG, let EEE be the edges of its DFS-Tree, then all EPCCEPCCEPCC are the SCCSCCSCC of G−EG - EG−E; therefore perform the SCC-Algorithm on the graph G−EG - EG−E; (understand the nature of SCC algorithm is very vital, EBCC,PBCCEBCC, PBCCEBCC,PBCC are both based on it)
+ cuz all edges are undirected, for a edge (cur−nex)(cur-nex)(cur−nex), nexnexnex must be in the Stack if nexnexnex has been visited by DFS, so the array On_stack can be removed.

例题

+ link

PBCC

Some properties of the Algorithm:
+ for a connected-graph GGG, directly perform the SCC-Algorithm on it (unlike EBCCEBCCEBCC which is acted on G−EG - EG−E not GGG), then every PBCCPBCCPBCC has the property that when we at dfs(cur) where cur is not DFS-root and if(false == vis[nex]){ dfs( nex); if( low_id[nex] == dfs_id[cur]){ @Loc_0}} satisfied (low_id[nex] must always ≤\leq≤ dfs_id[cur]); when we at @Loc_0, now all stack-points [O,cur,nex,...][O, cur, nex, ...][O,cur,nex,...] where [cur,nex,...][cur, nex, ...][cur,nex,...] forms a PBCCPBCCPBCC (cuz [root] - [cur] - [nex], so cur must be a Cut-Point;) now, you need pop out all nex, ... but keeping cur still in the stack;
__. one exception is cur == DFS_root, now it becomes [cur] - [nex] where cur maybe not a Cut-Point (cuz it has no Father-Point), but if it has at least two DFS-Unvisited-Points, i.e., [nex1]-[cur]-[nex] then cur is a Cut-Point;
+ cuz all edges are undirected, for a edge (cur−nex)(cur-nex)(cur−nex), nexnexnex must be in the Stack if nexnexnex has been visited by DFS, so the array On_stack can be removed.

涉及割点/PBCC时的解决思路

When facing a Cut-Point (PBCC) problem, you should follow the next steps:
+ if the target graph is unconnected, then divide it into all it CCCCCCs (cuz all CCCCCCs are independent to each other when handling Cut-Point), let GGG be one of its CCCCCC;
+ consider all PBCCPBCCPBCCs of GGG, rather than considering all CCCCCCs after deleting all Cut-Points from GGG;

例题

+ link
+ link

SCC

@NeedFinish, new analysis for SCC

过去分析

For a (directed) graph G=(V,E)G = (V,E)G=(V,E), let Id[v]Id[v]Id[v] is the SCC-ID of v∈Vv \in Vv∈V which within the range [0, SCC-Count).

A DFS-Order-Sequence of a graph is [0,1,2,...,n)[0, 1, 2, ..., n)[0,1,2,...,n), we use Order[v]Order[v]Order[v] represents its DFS-Order.

Let Low[v]Low[v]Low[v] denotes a DFS-Order such that Low[v]=Order[w]Low[v] = Order[w]Low[v]=Order[w] such that Order[w]≤Order[v]Order[w] \leq Order[v]Order[w]≤Order[v] and w,vw,vw,v are in the same SCC
. If there exists a point www such that Order[w]<Order[v]Order[w] < Order[v]Order[w]<Order[v] and w,vw,vw,v in the same SCC, then Low[v]<Order[v]Low[v] < Order[v]Low[v]<Order[v]; otherwise Low[v]=Order[v]Low[v] = Order[v]Low[v]=Order[v] if and only if Order[v]Order[v]Order[v] is the minimal-Order in its SCC (every SCC has exactly one such point, we call such a point The Root of SCC).

+ e.g., G=(1→2)(2→3)(3→4)(4→3)(2→5)(5→2)(2→1)G = (1\to 2)(2\to 3)(3\to 4)(4\to 3)(2\to 5)(5\to 2)(2\to 1)G=(1→2)(2→3)(3→4)(4→3)(2→5)(5→2)(2→1) (note the order of edges is also the process of DFS(1)), then we have 222 SCC (1,2,5)(3,4)(1,2,5) (3,4)(1,2,5)(3,4); finally, Low[1]=0,Low[2]=0,Low[5]=1Low[1] = 0, Low[2] = 0, Low[5]=1Low[1]=0,Low[2]=0,Low[5]=1, all Low[v]Low[v]Low[v] in a SCC maybe not the same; Low[2]=0Low[2] = 0Low[2]=0 cuz there is Order[1]=0<Order[2]=1Order[1] = 0 < Order[2]=1Order[1]=0<Order[2]=1; Low[5]=1Low[5] = 1Low[5]=1 cuz there is Order[2]=1<Order[5]=4Order[2] = 1 < Order[5]=4Order[2]=1<Order[5]=4 (Low[5]Low[5]Low[5] can also equals Order[1]=0Order[1] = 0Order[1]=0); 000 is the root of the SCC 1,2,51,2,51,2,5)

When we at Dfs(cur)Dfs( cur)Dfs(cur) and iterating its adjacent-points nexnexnex, there are several steps:
+ Firstly, we set Order[cur] = Low[cur] = DFS_order ++ (cuz Low[cur] <= Order[cur] finally, so we set it equals to Order[cur] initially)
+ If nex has never been called by DFS, so we call Dfs(nex)Dfs(nex)Dfs(nex); and then we update Low[cur]=min(self,Low[nex])Low[cur] = min(self, Low[nex])Low[cur]=min(self,Low[nex]) (cuz nex never be DFS visited, if cur,nexcur, nexcur,nex is not in the same SCC and Order[nex]>Order[cur]Order[nex] > Order[cur]Order[nex]>Order[cur], we have nexnexnex must the Root of its SCC, so Low[nex]>Order[cur]Low[nex] > Order[cur]Low[nex]>Order[cur]; this operation is ineffective), otherwise cur,nexcur,nexcur,nex in the same SCC, if curcurcur is the Root then Low[nex]=Order[cur]Low[nex] = Order[cur]Low[nex]=Order[cur], so this operations is also ineffective; if curcurcur is not the Root, there must exists a point nex1nex1nex1 whose Low[nex1]<Order[cur]Low[nex1] < Order[cur]Low[nex1]<Order[cur] (cuz there must has a path cur→nex1→..→rootcur \to nex1 \to .. \to rootcur→nex1→..→root where ......... do not contains curcurcur, therefore Low[nex1]=Order[root]<Order[cur]Low[nex1] = Order[root] < Order[cur]Low[nex1]=Order[root]<Order[cur]);

+ When DFS returns to a root-point Low[] = Order[], then the DFS-Stack must be in the form [...., root, S] where any point that in the same SCC with root, must occurs in S;

Code

void dfs( int _cur){queue[ tail ++] = _cur;isInQueue[ _cur] = true;dfsOrderId[ _cur] = dfsOrderId_counter ++;lowest_orderId[ _cur] = dfsOrderId[ _cur];//--for( int nex, i = graph->Head[ _cur]; ~i; i = graph->Next[ i]){nex = graph->Vertex[ i];if( -1 == dfsOrderId[ nex]){ //< never `dfs( nex)` calleddfs( nex);}if( isInQueue[ nex]){ //< `nex` maybe already forms a SCC, we should aviod this case; e.g., (1<->2)<-3, so Low[2] cann't be used to updated Low[3]lowest_orderId[ _cur] = min( lowest_orderId[ nex], lowest_orderId[ _cur]);}}if( lowest_orderId[ _cur] == dfsOrderId[ _cur]){scc_size[ scc_id_counter] = 0;while( true){int c = queue[ tail - 1];isInQueue[ c] = false;-- tail;scc_id[ c] = scc_id_counter;++ scc_size[ scc_id_counter];if( c == _cur){ break;}}++ scc_id_counter;}
}

Actually, the array dfsOrderId, lowest_orderId can be omitted; (use scc_id to denote lowest_orderId, cuz when DFS not yet returned to root, the array scc_id is free; and once a SCC is fixed (i.e., DFS has returned to its root, any other points (not belongs to this SCC) would not visit/use any information of this SCC)

 void dfs( int _cur){ptrNew_queue[ queue_tail ++] = _cur;ptrNew_isInQueue[ _cur] = true;int current_dfsOrderId = dfsOrderId_counter;ptrNew_sccId[ _cur] = dfsOrderId_counter ++;//--for( int nex, i = ptrRef_graph->Head[ _cur]; ~i; i = ptrRef_graph->Next[ i]){nex = ptrRef_graph->Vertex[ i];if( -1 == ptrNew_sccId[ nex]){ //< never `dfs( nex)` calleddfs( nex);}if( ptrNew_isInQueue[ nex]){ptrNew_sccId[ _cur] = min( ptrNew_sccId[ nex], ptrNew_sccId[ _cur]);}}if( ptrNew_sccId[ _cur] == current_dfsOrderId){ptrNew_sccSize[ scc_id_counter] = 0;while( true){int c = ptrNew_queue[ queue_tail - 1];ptrNew_isInQueue[ c] = false;-- queue_tail;ptrNew_sccId[ c] = scc_id_counter;++ ptrNew_sccSize[ scc_id_counter];if( c == _cur){ break;}}++ scc_id_counter;}
}

例题

+ link

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`Computer-Algorithm` Tarjan算法,强连通分量SCC,PBCC割点,EBCC割边/桥

Contents

EBCC

例题

PBCC

涉及割点/PBCC时的解决思路

例题

SCC

过去分析

例题

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