图论，求有向图的强连通分支

求有向图的强联通分支十一个典型的图论题目。
本来想把他写成一个通用的算法，但是发现有很多不太好处理的地方，比如选用邻接链表还是选用邻接矩阵来表示图，有的情况是不一样的，再者很难把强联通分支缩点，并且再构建一个图出来，主要是强联通分支之间的边表示的时候要考虑一些情况，比如边的费用什么的。
但是也可以尝试做一个稍微通用的一点的算法，就是把所有点属于那个强连通标识（用id[i]表示）出来。
如果对于原图中两点i,j。 如果id[i]==id[j]，则他们在一个强联通分支中，否则他们一定不在一个强联通分支中。至于确定两个不同强连通分支之间的关系，就留给客户去做，就好了。
先给出pku的一个题目，使用了求强联通分支的技术。

The Bottom of a Graph

Description
We will use the following (standard) definitions from graph theory. Let V be a nonempty and finite set, its elements being called vertices (or nodes). Let E be a subset of the Cartesian product V×V, its elements being called edges. Then G=(V,E) is called a directed graph.
Let n be a positive integer, and let p=(e₁,...,e_n) be a sequence of length n of edges e_i∈E such that e_i=(v_i,v_i+1) for a sequence of vertices (v₁,...,v_n+1). Then p is called a path from vertex v₁ to vertex v_n+1 in G and we say that v_n+1 is reachable from v₁, writing (v₁→v_n+1).
Here are some new definitions. A node v in a graph G=(V,E) is called a sink, if for every node w in G that is reachable from v, v is also reachable from w. The bottom of a graph is the subset of all nodes that are sinks, i.e., bottom(G)={v∈V|∀w∈V:(v→w)⇒(w→v)}. You have to calculate the bottom of certain graphs.

Input
The input contains several test cases, each of which corresponds to a directed graph G. Each test case starts with an integer number v, denoting the number of vertices of G=(V,E), where the vertices will be identified by the integer numbers in the set V={1,...,v}. You may assume that 1<=v<=5000. That is followed by a non-negative integer e and, thereafter, e pairs of vertex identifiers v₁,w₁,...,v_e,w_e with the meaning that (v_i,w_i)∈E. There are no edges other than specified by these pairs. The last test case is followed by a zero.

Output
For each test case output the bottom of the specified graph on a single line. To this end, print the numbers of all nodes that are sinks in sorted order separated by a single space character. If the bottom is empty, print an empty line.

Sample Input

3 3
1 3 2 3 3 1
2 1
1 2
0

Sample Output
1 3


import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.List;

public class Main
{

 static LinkedList [] mapa;
 static LinkedList [] mapt;
 static int [] order;
 static int tag;
 static boolean [] vst;
 static int [] id;
 static int [] id_degree_cnt;
 public static void main(String[] args) throws IOException
 {
  BufferedReader cin = new BufferedReader(new InputStreamReader(System.in));
  PrintWriter out = new PrintWriter(System.out);
  while(true)
  {
   int n=0,m=0;
   try
   {
    String s=cin.readLine().trim();
    String [] sa=s.split(" +");
    n=Integer.parseInt(sa[0]);
    if(n==0)break;
    m=Integer.parseInt(sa[1]);
    mapa=new LinkedList[n];
    mapt=new LinkedList[n];
    for(int i=0;i<n;i++)
    {
     mapa[i]=new LinkedList<Integer>();
     mapt[i]=new LinkedList<Integer>();
    }
    if(m>=0)
    {    
     s=cin.readLine().trim();
     sa=s.split(" +");
    }
    else sa=new String[] {};
    for(int i=0;i<sa.length;i+=2)
    {
     int from=Integer.parseInt(sa[i])-1;
     int to=Integer.parseInt(sa[i+1])-1;
     mapa[from].add(to);
     mapt[to].add(from);
    }
   }catch(Exception e)
   {
    
   }
   
   order=new int[n];
   tag=0;
   vst=new boolean[n];
   for(int i=0;i<n;i++)
    if(vst[i]==false)
    {
     vst[i]=true;
     dfsa(i);
    }
   id=new int[n];
   tag=0;
   Arrays.fill(vst,false);
   for(int i=n-1;i>=0;i--)
   {
    if(vst[order[i]]==false)
    {
     vst[order[i]]=true;
     dfst(order[i]);
     tag++;
    }
   }
   id_degree_cnt=new int[tag];
   for(int i=0;i<n;i++)
   {
    LinkedList ll=mapa[i];
    for(int j=0;j<ll.size();j++)
    {
     int next=(Integer)ll.get(j);
     if(id[i]!=id[next])
      id_degree_cnt[id[i]]++;
    }
   }
   List<Integer>rel=new ArrayList<Integer>(n);
   for(int i=0;i<n;i++)
    if(id_degree_cnt[id[i]]==0)
     rel.add(i);
   for(int d:rel)
    out.print((d+1)+" ");
   out.println();
   
  }
  out.flush();
  out.close();
 }
 public static void dfsa(int ci)
 {
 
  for(int i=0;i<mapa[ci].size();i++)
  {
   int next=(Integer)mapa[ci].get(i);
   if(vst[next]==false)
   {
    vst[next]=true;
    dfsa(next);
   }
  }
  order[tag++]=ci;
 }
 public static void dfst(int ci)
 {
  id[ci]=tag;
  for(int i=0;i<mapt[ci].size();i++)
  {
   int next=(Integer)mapt[ci].get(i);
   if(vst[next]==false)
   {
    vst[next]=true;
    dfst(next);
   }
  }
 }
 
}
2