POJ 2553 The Bottom of a Graph

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 10687		Accepted: 4403

Description

We'll use the following-definitions from graph theory. Let VBe a nonempty and finite set, its elements being called vertices (or nodes). Let EBe a subset of the Cartesian product VXV, its elements being called edges. Then g= (v,e)is called a directed graph.
Let NBe a positive integers, and let p= (E1,..., en)Be a sequence of length Nof edges ei∈esuch that ei= (vi,vi+1)For a sequence of vertices (v1,..., vn+1). Then Pis called a path from vertex v1to vertex vn+1Inch GAnd we say that vn+1is reachable from v1, writing (v1→vn+1).
Here is some new definitions. A node vIn a graph g= (v,e)is called a sink, if for every node WInch GThat's reachable from v, vis also reachable from W. The bottom of a graph is the subset of all nodes it is 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 the 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 is identified by the integer numbers in the set v={1,..., V}. Assume that 1<=v<=5000. That's followed by a non-negative integer eAnd, thereafter, ePairs of vertex identifiers v1,w1,..., ve,weWith the meaning. (VI,WI) ∈e. There is no edges other than specified by these pairs. The last test case was followed by a zero.

Output

For all test case output the bottom of the specified graph on a single line. To this end, print the numbers of any nodes that is sinks in sorted order separated by a single space character. If the bottom is empty, print a empty line.

Sample Input

3 31 3 2 3 3 12 11 20

Sample Output

1 32

Source

ULM Local 2003

1#include <iostream>2#include <cstring>3#include <cstdio>4#include <vector>5#include <algorithm>6#include <stack>7 #defineMAXM 5000108 #defineMAXN 50109 using namespacestd;Ten intt,n,m; One structedge{ A     intU,v,next; - }E[MAXM],EE[MAXM]; -vector<int>Kp[maxn],ans; the intHEAD[MAXN],JS,HEADD[MAXN],JSS; - BOOLEXIST[MAXN]; - intVisx,cur;//cur--the number of points to indent - intDFN[MAXN],LOW[MAXN],BELONG[MAXN],CHUDU[MAXN]; +stack<int>St; - voidinit () { +memset (Chudu,0,sizeof(Chudu)); AMemset (E,0,sizeof(e)); memset (EE,0,sizeof(EE)); atMemset (Head,0,sizeof(head)); memset (Headd,0,sizeof(Headd)); -Jss=js=visx=cur=0; -memset (exist,false,sizeof(exist)); -      while(!st.empty ()) St.pop (); -memset (DFN,0,sizeof(DFN)); memset (Low,0,sizeof(Low)); -Memset (Belong,0,sizeof(belong)); inAns.resize (0); -      for(intI=0; i<maxn;i++) Kp[i].resize (0); to } + voidAdd_edge1 (intUintv) { -e[++js].u=u;e[js].v=v; thee[js].next=head[u];head[u]=JS; * } $ voidTarjan (intu) {Panax Notoginsengdfn[u]=low[u]=++Visx; -exist[u]=true; the st.push (u); +      for(intI=head[u];i;i=E[i].next) { A         intv=e[i].v; the         if(dfn[v]==0){ + Tarjan (v); -low[u]=min (low[u],low[v]); $         } $         Else if(Exist[v]&&low[u]>dfn[v]) low[u]=Dfn[v]; -     } -     intJ; the     if(low[u]==Dfn[u]) { -++cur;Wuyi          Do{ theJ=st.top (); St.pop (); exist[j]=false; - Kp[cur].push_back (j); Wubelong[j]=cur; -} while(j!=u); About     } $ } - voidAdd_edge2 (intUintv) { -ee[++jss].u=u;ee[jss].v=v; -ee[jss].next=headd[u];headd[u]=JSS; A } + voidApki (intx) { the      for(intI=0; I<kp[x].size (); i++){ -         intk=Kp[x][i]; $ Ans.push_back (k); the     } the } the intMain () the { -      while(SCANF ("%d%d", &n,&m) = =2){ in init (); the         intu,v; the          for(intI=0; i<m;i++){ Aboutscanf"%d%d",&u,&v); Add_edge1 (u,v); the         } the          for(intI=1; i<=n;i++) {//finding strong connected components the             if(dfn[i]==0) Tarjan (i); +         } -          for(intI=1; i<=m;i++){ the             intu=e[i].u,v=e[i].v;Bayi             if(belong[u]!=Belong[v]) { the Add_edge2 (Belong[u],belong[v]); thechudu[belong[u]]++; -             } -         } the          for(intI=1; i<=cur;i++){ the             if(chudu[i]==0) the Apki (i); the         } - sort (Ans.begin (), Ans.end ()); the          for(intI=0; I<ans.size (); i++) printf ("%d", Ans[i]); theprintf"\ n"); the     }94     return 0; the}

Thought: And POJ2762 not much difference

POJ 2553 The Bottom of a Graph