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;
}

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