【POJ】The Bottom of a Graph 强连通

The Bottom of a Graph

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 8636		Accepted: 3582

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
2

Source

Ulm Local 2003

传送门：【POJ】The Bottom of a Graph

题目大意：给一个有向图，问出度为0的点的集合（将相互可达的点看成一个点，如果该点是出度为0，则该点所包括的所有点都算出度为0的点）。

题目分析：强连通缩点。然后对所有点按编号从小到大扫描，如果所属的分量出度为0，输出。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define clear( a , x ) memset ( a , x , sizeof a )

const int MAXN = 5001 ;
const int MAXE = 100000 ;

struct Edge {
	int v , n ;
	Edge ( int var = 0 , int next = 0 ) : v(var) , n(next) {}
} ;

struct SCC {
	Edge edge[MAXE] ;
	int adj[MAXN] , cntE ;
	int Dfn[MAXN] , Low[MAXN] , dfs_clock ;
	int scc[MAXN] , scc_cnt ;
	int S[MAXN] , top ;
	bool ins[MAXN] ;
	bool ou[MAXN] ;
	
	void init () {
		top = 0 ;
		cntE = 0 ;
		scc_cnt = 0 ;
		dfs_clock = 0 ;
		clear ( ou , 0 ) ;
		clear ( ins , 0 ) ;
		clear ( Dfn , 0 ) ;
		clear ( adj , -1 ) ;
	}
	
	void addedge ( int u , int v ) {
		edge[cntE] = Edge ( v , adj[u] ) ;
		adj[u] = cntE ++ ;
	}
	
	void Tarjan ( int u ) {
		Dfn[u] = Low[u] = ++ dfs_clock ;
		S[top ++] = u ;
		ins[u] = 1 ;
		for ( int i = adj[u] ; ~i ; i = edge[i].n ) {
			int v = edge[i].v ;
			if ( !Dfn[v] ) {
				Tarjan ( v ) ;
				Low[u] = min ( Low[u] , Low[v] ) ;
			}
			else if ( ins[v] )
				Low[u] = min ( Low[u] , Dfn[v] ) ;
		}
		if ( Low[u] == Dfn[u] ) {
			++ scc_cnt ;
			while ( 1 ) {
				int v = S[-- top] ;
				ins[v] = 0 ;
				scc[v] = scc_cnt ;
				if ( v == u )
					break ;
			}
		}
	}
	
	void find_scc ( int n ) {
		REPF ( i , 1 , n )
			if ( !Dfn[i] )
				Tarjan ( i ) ;
	}
	
	void solve ( int n ) {
		REPF ( u , 1 , n )
			for ( int i = adj[u] ; ~i ; i = edge[i].n ) {
				int v = edge[i].v ;
				if ( scc[u] != scc[v] ) {
					ou[scc[u]] = 1 ;
				}
			}
		int flag = 0 ;
		REPF ( i , 1 , n )
			if ( !ou[scc[i]] ) {
				if ( flag )
					printf ( " " ) ;
				flag = 1 ;
				printf ( "%d" , i ) ;
			}
		printf ( "\n" ) ;
	}
} ;

SCC C ;

void work () {
	int n , m ;
	int u , v ;
	while ( ~scanf ( "%d%d" , &n , &m ) && n ) {
		C.init () ;
		while ( m -- ) {
			scanf ( "%d%d" , &u , &v ) ;
			C.addedge ( u , v ) ;
		}
		C.find_scc ( n ) ;
		C.solve ( n ) ;
	}
}

int main () {
	work () ;
	return 0 ;
}