POJ2553解题报告 强连通分支
The Bottom of a Graph
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 4000 | Accepted: 1637 |
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.
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
我的tarjan第二题
题目所求：按升序输出出度为0的强连通分支所包含的点。
思路：先用tarjan算法标记各个强连通分支内的点，然后求出每个连通分支的出度，找出出度为0的强连通分支并标记该分支内的点，最后
按照顺序输出符合条件的每个点。。
#include<iostream>
using namespace std;
int S[6000],C,dfn[6000],low[6000],stack[6000],top,t;
bool in[6000];
struct L
{
int v; L *next;
};
L *head[6000];
void tarjan(int v)
{
dfn[v]=low[v]=++t;
stack[top++]=v;
in[v]=true;
for(L *p=head[v];p!=NULL;p=p->next)
if(!dfn[p->v])
{
tarjan(p->v);
if(low[p->v]<low[v])
low[v]=low[p->v];
}
else if(in[p->v]&&low[p->v]<low[v])
low[v]=low[p->v];
if(dfn[v]==low[v])
{
C++;
do
{
v=stack[--top];
S[v]=C;
in[v]=false;
}while(dfn[v]!=low[v]);
}
}
int main()
{
int n,m,i,a,b;
while(cin>>n&&n&&cin>>m)
{
memset(head,0,sizeof(int)*6000);
memset(S,0,sizeof(int)*6000);
memset(dfn,0,sizeof(int)*6000);
memset(low,0,sizeof(int)*6000);
memset(in,0,sizeof(bool)*6000);
for(i=0;i<m;i++)
{
cin>>a>>b;
L *k=new L;
k->next=head[a];
head[a]=k;
k->v=b;
}
C=top=t=0;
for(i=1;i<=n;i++)
if(!dfn[i])
tarjan(i);
int chudu[6000]={0}; bool O[6000]={false};
for(i=1;i<=n;i++)
for(L *p=head[i];p!=NULL;p=p->next)
if(S[i]!=S[p->v])
chudu[S[i]]++;
for(i=1;i<=C;i++)
if(!chudu[i])
{
for(a=1;a<=n;a++)
if(S[a]==i)
O[a]=true;
}
for(a=1;a<=n;a++)
if(O[a])
cout<<a<<' ';
cout<<endl;
}
return 0;
}