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

Sample Output

1 3

2

Source

Ulm Local 2003

题意:使用的图论的方式说明了一个新的定义,汇点的定义,v是图中的一个顶点,对于图中的每一个顶点w,如果v可达w并且w也可达v,ze称v为汇点。图的底部为图的子集,子集中的所有的点都是汇点,求图的底部。

思路:如果图的底部都是汇点,则说明底部中的任意两点都互相可达,则底部为强连通分量,并且这个集合不与外部相连即从这个集合不能到达其他的集合,所以任务就变成求图的强连通分量并且出度为零

#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 Head[Max],top; 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; void AddEdge(int u,int v)
{
Li[top].v = v; Li[top].next = Head[u]; Head[u] = top++;
} void Tarjan(int u) // Tarjan求强连通分量
{ dfn[u]=low[u]=dep++; S.push(u); for(int i=Head[u];i!=-1;i=Li[i].next)
{
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; memset(Head,-1,sizeof(Head)); for(int i=0;i<m;i++)
{
scanf("%d %d",&u,&v); AddEdge(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++)
{
for(int j=Head[G[i][k]];j!=-1;j = Li[j].next)
{
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;
}