有向图的强连通算法 -- tarjan算法

（绘图什么真辛苦）

强连通分量：

在有向图 G 中。若两个顶点相互可达，则称两个顶点强连通(strongly
connected)。

假设有向图G的每两个顶点都强连通，称G是一个强连通图。非强连通图有向图的极大强连通子图。称为强连通分量(strongly
connected components)。

比方上面这幅图(
a, b, e ), ( d, c, h ), ( f, g ) 分别为三个 SCC。

tarjan算法伪代码：

该算法由 Robert
Tarjan 发明，原论文：Tarjan1972

时间复杂度是深搜的时间复杂度 O( N
+ E )。

BEGIN
    INTERGER i;
    PROCEDURE STRONGCONNECT(v);
        BEGIN
            LOWLINK(v):= NUMBER(v):= i := i + 1;
            put v on stack of points;
            FOR w in the adjacency list of v DO
                BEGIN
                    IF w is not yet numbered THEN
                        BEGIN comment( v, w ) is a tree arc;
                            STRONGCONNECT(w);
                            LOWLINK(V) := min( LOWLINK(V),
                                               LOWLINK(W));
                        END;
                    ELSE IF NUMBER(W) < NUMBER(V) DO
                        BEGIN comment( v, w ) is a frond or cross-link;
                            if w is on stack of points THEN
                                LOWLINK(v) := min( LOWLINK(v),
                                                   NUMBER(w));
                        END;
                END;

            if( LOWLINK(v) = NUMBER(v) ) THEN
                BEGIN comment v is the root of a compont;
                    start new strongly connected compont;
                    WHILE w on top of point stack satisfies
                        NUMBER(w) >= NUMBER(v) DO
                        BEGIN
                                delete w from point stack and put w
                                in current component;
                        END;
                END;
        END;
    i := 0;
    empty stack of points;
    FOR w a vertex IF w is not yet numbered THEN STRONGCONNECT(w);
END

tarjan算法的运行动态图：

1.建图。每一个顶点有三个域。第一个是顶点名。第二个空格是发现该点的时间戳 DFN ，第三个空格是该点可以追溯到的最早的栈中节点的次序号
LOW。

2. 深搜，比方搜索路径为 a --> b --> c --> g --> f。 沿途记下自身的 DFN 和 LOW

3.达到 f 点后，f 点仅仅有一个可达顶点 g， 但 g 不是 f 的后继顶点。且 g 在栈中，则更新 f 的 LOW 变为 g 的发现时间。

4.这时候回到 g 点。可是 f 点还得再栈中，仅仅有发现时间 DFN 与 LOW 同样的顶点才干从栈中弹出（和压在上面的节点一起弹出构成SCC）

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5.这时候 g 点的 DFN == LOW

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6.于是将 f 和 g 都弹出

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7.如虚线所看到的，他们构成了一个SCC

8.以下就是一样的了，（ 图画的好辛苦 ）

     

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python代码：

def strongly_connected_components( graph ):

    dfn_count = [0]
    result = []
    stack = []
    low = {}
    dfn = {}

    def stronglyconnected( node ):

        dfn[node] = dfn_count[0]
        low[node] = dfn_count[0]
        dfn_count[0] += 1
        stack.append( node )

        if node not in graph:
            successors = []
        else:
            successors = graph[node]

        for successor in successors:
            if successor not in dfn:
                stronglyconnected( successor )
                low[node] = min( low[successor], low[node] )
            elif successor in stack:
                low[node] = min( low[node], dfn[successor] )

        if low[node] == dfn[node]:
            li = []
            item = None
            while True:
                item = stack.pop()
                li.append( item )
                if item == node: break
            result.append( li )

    for node in graph:
        if node not in low:
            stronglyconnected( node )

    return result

if __name__ == '__main__':
    graph = {
        'a': [ 'b' ],
        'b': [ 'c', 'e', 'f' ],
        'c': [ 'g', 'd' ],
        'd': [ 'c', 'h' ],
        'e': [ 'a', 'f' ],
        'f': [ 'g' ],
        'g': [ 'f' ],
        'h': [ 'g', 'd' ]
    }
    print strongly_connected_components( graph )

执行结果：

C++代码：

#include <iostream>
#include <vector>
#include <string.h>
using namespace std;

#define MAX_SIZE 100

bool Graph[MAX_SIZE][MAX_SIZE];

int Stack[MAX_SIZE];
bool nodeIsInStack[MAX_SIZE];
int stack_pointer = 0;

int Lows[MAX_SIZE];
int Dfns[MAX_SIZE];

int node_num = 0;   // 图中节点得到个数
int find_time = 0;  // 每一个节点的发现时间

int scc_num = 0;    // 记录 scc 的个数

void find_scc( int start_node ){

    find_time++;
    Lows[start_node] = Dfns[start_node] = find_time;

    stack_pointer++;
    Stack[stack_pointer] = start_node;
    nodeIsInStack[start_node] = true;

    for( int end_node = 1; end_node <= node_num; ++end_node ){
        //若start_node和end_ndoe之间存在边
        if( Graph[start_node][end_node] ){
            //若是end_node尚未被訪问
            if( Dfns[end_node] == 0 ){
                find_scc( end_node );
                Lows[start_node] = min( Lows[start_node], Lows[end_node] );
            }
            //若end_node在栈中。也就是start_node -> end_node是返祖边
            else if( nodeIsInStack[end_node] ){
                Lows[start_node] = min( Lows[start_node], Dfns[end_node] );
            }
        }
    }

    //若是start_node的时间戳与Lows相等在构成SCC
    if( Dfns[start_node] == Lows[start_node] ){

        scc_num++;
        cout << "scc_num: " << scc_num << endl;

        int pop_node_index = Stack[stack_pointer];
        stack_pointer--;
        nodeIsInStack[pop_node_index] = false;

        cout << pop_node_index << " ";

        while( start_node != pop_node_index ){

            pop_node_index = Stack[stack_pointer];
            nodeIsInStack[pop_node_index] = false;
            stack_pointer--;
            cout << pop_node_index << " ";
        }
        cout << endl;
    }
}

void init_values(){
    memset( Graph, false, sizeof( Graph ) );
    memset( Stack, 0, sizeof( Stack ) );
    memset( nodeIsInStack, false, sizeof( nodeIsInStack ) );
    memset( Lows, 0, sizeof( Lows ) );
    memset( Dfns, 0, sizeof( Dfns ) );
}

int main(){
    init_values();
    // 初始化图
    //这里用数字取代节点的名字
    int start_node, end_node;
    cin >> node_num;

    while( true ){
        cin >> start_node >> end_node;
        //起始点终止点都为 0 的时候结束
        if( start_node == 0 && end_node == 0 )
            break;
        Graph[start_node][end_node] = true;
    }

    for( int start_node = 1; start_node <= node_num; ++start_node ){
        //该节点尚未被訪问到
        if( Dfns[start_node] == 0 ){
            find_scc( start_node );
        }
    }

}