TarJan 算法求解有向连通图强连通分量

tarjan(u)
{

    DFN[u]=Low[u]=++Index     // 为节点u设定次序编号和Low初值

    Stack.push(u)                     // 将节点u压入栈中

    for each (u, v) in E               // 枚举每一条边

          if (v is not visted)          // 如果节点v未被访问过

                  tarjan(v)              // 继续向下找

                  Low[u] = min(Low[u], Low[v])

            else if (v in S)            // 如果节点v还在栈内

            Low[u] = min(Low[u], DFN[v])

    if (DFN[u] == Low[u])        // 如果节点u是强连通分量的根

       repeat

           v = S.pop                  // 将v退栈，为该强连通分量中一个顶点

           print v

      until (u== v)

} 

#include "cstdlib"
#include "cctype"
#include "cstring"
#include "cstdio"
#include "cmath"
#include "algorithm"
#include "vector"
#include "string"
#include "iostream"
#include "sstream"
#include "set"
#include "queue"
#include "stack"
#include "fstream"
#include "strstream"
using namespace std;

#define  M 2000              //题目中可能的最大点数      
int STACK[M],top=0;          //Tarjan 算法中的栈
bool InStack[M];             //检查是否在栈中
int DFN[M];                  //深度优先搜索访问次序
int Low[M];                  //能追溯到的最早的次序
int ComponetNumber=0;        //有向图强连通分量个数
int Index=0;                 //索引号
vector <int> Edge[M];        //邻接表表示
vector <int> Component[M];   //获得强连通分量结果

void Tarjan(int i)
{
    int j;
    DFN[i]=Low[i]=Index++;
    InStack[i]=true;
    STACK[++top]=i;
    for (int e=0;e<Edge[i].size();e++)
    {
        j=Edge[i][e];
        if (DFN[j]==-1)
        {
            Tarjan(j);
            Low[i]=min(Low[i],Low[j]);
        }
        else if (InStack[j])
            Low[i]=min(Low[i],DFN[j]);
    }
    if (DFN[i]==Low[i])
    {
        cout<<"TT    "<<i<<"   "<<Low[i]<<endl;
        ComponetNumber++;
        do
        {
            j=STACK[top--];
            InStack[j]=false;
            Component[ComponetNumber].push_back(j);
        }
        while (j!=i);
    }
}

void solve(int N)     //此图中点的个数，注意是0-indexed！
{
    memset(STACK,-1,sizeof(STACK));
    memset(InStack,0,sizeof(InStack));
    memset(DFN,-1,sizeof(DFN));
    memset(Low,-1,sizeof(Low));
    for(int i=0;i<N;i++)
        if(DFN[i]==-1)
            Tarjan(i);   
}
/*
此算法正常工作的基础是图是0-indexed的。
*/
int main()
{
    Edge[0].push_back(1);Edge[0].push_back(2);
    Edge[1].push_back(3);
    Edge[2].push_back(3);Edge[2].push_back(4);
    Edge[3].push_back(0);Edge[3].push_back(5);
    Edge[4].push_back(5);
    int  N=6;
    solve(N);
    cout<<"ComponetNumber is "<<ComponetNumber<<endl;
    for(int i=0;i<N;i++)
        cout<<Low[i]<<" ";
    cout<<endl;
    for(int i=0;i<N;i++)
    {
        for(int j=0;j<Component[i].size();j++)
            cout<<Component[i][j];
        cout<<endl;
    }
    return 0;
}