poj--2553--The Bottom of a Graph (scc+缩点）

The Bottom of a Graph

Time Limit : 6000/3000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 1   Accepted Submission(s) : 1

Problem 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
#include<stdio.h>
#include<string.h>
#include<queue>
#include<stack>
#include<algorithm>
#include<vector>
using namespace std;
#define MAX 50010
struct node
{
	int u,v;
	int next;
}edge[MAX];
int low[MAX],dfn[MAX];
int sccno[MAX],head[MAX];
int scc_cnt,dfs_clock,cnt;
bool Instack[MAX];
int m,n;
stack<int>s;
vector<int>G[MAX];
vector<int>scc[MAX];
int in[MAX],out[MAX];
int num[MAX];
void init()
{
	memset(head,-1,sizeof(head));
	cnt=0;
}
void add(int u,int v)
{
	edge[cnt].u=u;
	edge[cnt].v=v;
	edge[cnt].next=head[u];
	head[u]=cnt++;
}
void getmap()
{
	int a,b;
	while(m--)
	{
		scanf("%d%d",&a,&b);
		add(a,b);
	}
}
void tarjan(int u,int fa)
{
	int v;
	low[u]=dfn[u]=++dfs_clock;
	s.push(u);
	Instack[u]=true;
	for(int i=head[u];i!=-1;i=edge[i].next)
	{
		v=edge[i].v;
		if(!dfn[v])
		{
			tarjan(v,u);
			low[u]=min(low[u],low[v]);
		}
		else if(Instack[v])
		low[u]=min(low[u],dfn[v]);
	}
	if(low[u]==dfn[u])
	{
		scc_cnt++;
		scc[scc_cnt].clear();
		for(;;)
		{
			v=s.top();
			s.pop();
			Instack[v]=false;
			scc[scc_cnt].push_back(v);
			sccno[v]=scc_cnt;
			if(v==u) break;
		}
	}
}
void find(int l,int r)
{
	memset(sccno,0,sizeof(sccno));
	memset(low,0,sizeof(low));
	memset(dfn,0,sizeof(dfn));
	memset(Instack,false,sizeof(Instack));
	dfs_clock=scc_cnt=0;
	for(int i=l;i<=r;i++)
	if(!dfn[i])
	tarjan(i,-1);
}
void suodian()
{
	for(int i=1;i<=scc_cnt;i++)
	G[i].clear(),in[i]=out[i]=0;
	for(int i=0;i<cnt;i++)
	{
		int u=sccno[edge[i].u];
		int v=sccno[edge[i].v];
		if(u!=v)
		G[u].push_back(v),out[u]++,in[v]++;
	}
}
void solve()
{
	int ans=0;
	int k=0;
	for(int i=1;i<=scc_cnt;i++)
	{
		if(out[i]==0)
		{
			for(int j=0;j<scc[i].size();j++)
			num[k++]=scc[i][j];
		}
	}
	sort(num,num+k);
	for(int i=0;i<k-1;i++)
	printf("%d ",num[i]);
	printf("%d\n",num[k-1]);
}
int main()
{
	while(scanf("%d%d",&n,&m),n)
	{
		init();
		getmap();
		find(1,n);
		suodian();
		solve();
	}
	return 0;
}