POJ——T2553 The Bottom of a Graph

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

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 10987		Accepted: 4516

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. ThenG=(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+1is 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

 

模板题，调了三天（崩溃）很无奈

 

 #include <algorithm>
 #include <cstring>
 #include <cstdio>

 using namespace std;

 const int MAXN();
 const int N();
 int n,m,u,v;
 int sumedge,head[N];
 struct Edge
 {
     int to,next;
     Edge(int to=,int next=) :
         to(to),next(next) {}
 }edge[MAXN];

 void ins(int from,int to)
 {
     edge[++sumedge]=Edge(to,head[from]);
     head[from]=sumedge;
 }

 int low[N],dfn[N],tim;
 int Stack[N],instack[N],top;
 int sumcol,col[N],point[N];

 void DFS(int now)
 {
     low[now]=dfn[now]=++tim;
     Stack[++top]=now; instack[now]=true;
     for(int i=head[now];i;i=edge[i].next)
     {
         int go=edge[i].to;
         if(!dfn[go])
                 DFS(go),low[now]=min(low[now],low[go]);
         else if(instack[go]) low[now]=min(low[now],dfn[go]);
     }
     if(low[now]==dfn[now])
     {
         col[now]=++sumcol;
         for(;Stack[top]!=now;top--)
         {
             col[Stack[top]]=sumcol;
             instack[Stack[top]]=false;
         }
         instack[now]=false; top--;
     }
 }

 int cnt,ans[N],chudu[N];

 void init()
 {
     tim=top=cnt=;
     sumcol=sumedge=;
     memset(ans,,sizeof(ans));
     memset(low,,sizeof(low));
     memset(dfn,,sizeof(dfn));
     memset(col,,sizeof(col));
     memset(head,,sizeof(head));
     memset(chudu,,sizeof(chudu));
     memset(Stack,,sizeof(Stack));
     memset(instack,,sizeof(instack));
 }

 int main()
 {
     /*freopen("made.txt","r",stdin);
     freopen("myout.txt","w",stdout);*/

     while(~scanf("%d",&n)&&n)
     {
         scanf("%d",&m); init();
         for(;m;m--)
             scanf("%d%d",&u,&v),ins(u,v);
         for(int i=;i<=n;i++)
             if(!dfn[i]) DFS(i);
         for(u=;u<=n;u++)
             for(v=head[u];v;v=edge[v].next)
                 if(col[u]!=col[edge[v].to]) chudu[col[u]]++;
         /*for(int sc=1;sc<=sumcol;sc++)
           if(!chudu[sc])
             for(int i=1;i<=n;i++)
             if(sc==col[i]) ans[++cnt]=i;*/
         for(int i=;i<=n;i++)
             if(!chudu[col[i]]) ans[++cnt]=i;
         sort(ans+,ans+cnt+);
         if(cnt)
         {
             for(int i=;i<cnt;i++) printf("%d ",ans[i]);
             printf("%d\n",ans[cnt]);
         }
         else printf("\n");
     }
     return ;
 }