### The Bottom of a Graph

Time Limit: 3000MS Memory Limit: 65536K

Total Submissions: 9759 Accepted: 4053

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,…,en) be a sequence of length n of edges ei∈E such that ei=(vi,vi+1) for a sequence of vertices (v1,…,vn+1). Then p is called a path from vertex v1 to vertex vn+1 in G and we say that vn+1 is reachable from v1, writing (v1→vn+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,…,ve,we with the meaning that (vi,wi)∈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

1 3

2

##### Source

Ulm Local 2003

``````#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <set>
#include <vector>
#include <algorithm>

using namespace std;

const int Max = 5010;

typedef struct node
{
int v;

int next;
}Line;

Line Li[Max*1000];

int dfn[Max],low[Max],pre[Max],dep;

vector<int>G[Max];

int a[Max],num,Du[Max],Num;

bool vis[Max];

stack <int> S;

int n,m;

{
Li[top].v = v; Li[top].next = Head[u];

}

void Tarjan(int u) // Tarjan求强连通分量
{

dfn[u]=low[u]=dep++;

S.push(u);

{
if(dfn[Li[i].v]==-1)
{
Tarjan(Li[i].v);

low[u] = min(low[u],low[Li[i].v]);
}
else
{
low[u]=min(low[u],dfn[Li[i].v]);
}
}

if(low[u]==dfn[u])// 如果low[u]=dfn[u],则说明是强连通分的根节点
{
while(!S.empty())
{
int v = S.top();

S.pop();

G[Num].push_back(v);

pre[v]=Num;

if(v==u)
{
break;
}
}

Num++;
}
}

int main()
{
int u, v;

while(~scanf("%d",&n)&&n)
{
scanf("%d",&m);

top = 0;

for(int i=0;i<m;i++)
{
scanf("%d %d",&u,&v);

}

memset(dfn,-1,sizeof(dfn));

for(int i=0;i<=n;i++)
{
G[i].clear();
}

dep = 0;Num = 0;

for(int i=1;i<=n;i++)
{
if(dfn[i]==-1)
{
Tarjan(i);
}
}

memset(Du,0,sizeof(Du));

for(int i=0;i<Num;i++)
{
memset(vis,false,sizeof(vis));

for(int k=0;k<G[i].size();k++)
{
{
if(i != pre[Li[j].v]&&!vis[pre[Li[j].v]])//集合间度的计算
{
vis[pre[Li[j].v]]=true;

Du[i]++;
}
}
}
}

num = 0;

for(int i=0;i<Num;i++)
{
if(Du[i]==0)
{
for(int j=0;j<G[i].size();j++)
{
a[num++]=G[i][j];
}
}
}
sort(a,a+num);// 排序输出

for(int i=0;i<num;i++)
{
if(i)
{
printf(" ");
}
printf("%d",a[i]);
}
printf("\n");

}

return 0;
}``````