[tarjan] poj 2553 The Bottom of a Graph

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本文介绍了一道名为“图底”的POJ 2553竞赛题目，该题旨在寻找有向图中的一类特殊节点——sink节点。文章详细解释了如何使用Tarjan算法结合强连通分量缩点的方法来解决这个问题，并提供了完整的代码实现。

题目链接：

http://poj.org/problem?id=2553

The Bottom of a Graph

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 8899		Accepted: 3686

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

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题目意思：

给一幅有向图，求这样的节点i的集合,使得如果i到其他任意节点j可达的话，j也有路劲可达i。

解题思路：

tarjan+缩点+统计出度为0的强连通分量，并输出其中的节点。

代码：

//#include<CSpreadSheet.h>

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;


#define Maxn 5500
int low[Maxn],dfn[Maxn],sta[Maxn],n,m;
int dep,bc,sc,in[Maxn];
bool iss[Maxn];
vector<vector<int> >myv;
int de[Maxn];

void tarjan(int cur)
{
    int ne;

    low[cur]=dfn[cur]=++dep;
    sta[++sc]=cur;
    iss[cur]=true;

    for(int i=0;i<myv[cur].size();i++)
    {
        ne=myv[cur][i];
        if(!dfn[ne])
        {
            tarjan(ne);
            low[cur]=min(low[cur],low[ne]);
        }
        else if(iss[ne]&&dfn[ne]<low[cur])
            low[cur]=dfn[ne];
    }

    if(low[cur]==dfn[cur])
    {
        bc++;
        do
        {
            ne=sta[sc--];
            in[ne]=bc;
            iss[ne]=false;
        }while(ne!=cur);
    }

}
void solve()
{
    sc=bc=dep=0;
    memset(iss,false,sizeof(iss));
    memset(dfn,0,sizeof(dfn));
    for(int i=1;i<=n;i++)
        if(!dfn[i])
            tarjan(i);
}
int main()
{
    //freopen("in.txt","r",stdin);
   //freopen("out.txt","w",stdout);

   while(scanf("%d",&n)&&n)
   {
       scanf("%d",&m);
       myv.clear();
       myv.resize(n+1);

       for(int i=1;i<=m;i++)
       {
           int a,b;
           scanf("%d%d",&a,&b);
           myv[a].push_back(b);
       }
       solve();
       memset(de,0,sizeof(de));
       for(int i=1;i<=n;i++)
       {
           for(int j=0;j<myv[i].size();j++)
           {
               int ne=myv[i][j];

               if(in[i]!=in[ne])
                    de[in[i]]++;
           }
       }
       bool fi=true;
       for(int i=1;i<=n;i++)
       {
           if(!de[in[i]])
           {
               if(!fi)
                    putchar(' ');
               else
                    fi=false;
               printf("%d",i);
           }
       }
       printf("\n");
   }
    return 0;
}