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POJ——T2553 The Bottom of a Graph

2024-09-23 15:17:22   204人阅读

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,...,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+1is 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.技术分享

Sample Input

3 31 3 2 3 3 12 11 20

Sample Output

1 32

Source

Ulm Local 2003
 
模板题，调了三天（崩溃）很无奈
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 1 #include <algorithm> 2 #include <cstring> 3 #include <cstdio> 4  5 using namespace std; 6  7 const int MAXN(500010); 8 const int N(5010); 9 int n,m,u,v;10 int sumedge,head[N];11 struct Edge12 {13     int to,next;14     Edge(int to=0,int next=0) :15         to(to),next(next) {}16 }edge[MAXN];17 18 void ins(int from,int to)19 {20     edge[++sumedge]=Edge(to,head[from]);21     head[from]=sumedge;22 }23 24 int low[N],dfn[N],tim;25 int Stack[N],instack[N],top;26 int sumcol,col[N],point[N];27 28 void DFS(int now)29 {30     low[now]=dfn[now]=++tim;31     Stack[++top]=now; instack[now]=true;32     for(int i=head[now];i;i=edge[i].next)33     {34         int go=edge[i].to;35         if(!dfn[go])36                 DFS(go),low[now]=min(low[now],low[go]);37         else if(instack[go]) low[now]=min(low[now],dfn[go]);38     }39     if(low[now]==dfn[now])40     {41         col[now]=++sumcol;42         for(;Stack[top]!=now;top--)43         {44             col[Stack[top]]=sumcol;45             instack[Stack[top]]=false;46         }47         instack[now]=false; top--;48     }49 }50 51 int cnt,ans[N],chudu[N];52 53 void init()54 {55     tim=top=cnt=0;56     sumcol=sumedge=0;57     memset(ans,0,sizeof(ans));58     memset(low,0,sizeof(low));59     memset(dfn,0,sizeof(dfn));60     memset(col,0,sizeof(col)); 61     memset(head,0,sizeof(head)); 62     memset(chudu,0,sizeof(chudu));63     memset(Stack,0,sizeof(Stack)); 64     memset(instack,0,sizeof(instack));65 }66 67 int main()68 {69     /*freopen("made.txt","r",stdin);70     freopen("myout.txt","w",stdout);*/71 72     while(~scanf("%d",&n)&&n)73     {74         scanf("%d",&m); init();75         for(;m;m--)76             scanf("%d%d",&u,&v),ins(u,v);      77         for(int i=1;i<=n;i++)78             if(!dfn[i]) DFS(i);79         for(u=1;u<=n;u++)80             for(v=head[u];v;v=edge[v].next)81                 if(col[u]!=col[edge[v].to]) chudu[col[u]]++;82         /*for(int sc=1;sc<=sumcol;sc++)83           if(!chudu[sc])84             for(int i=1;i<=n;i++)85             if(sc==col[i]) ans[++cnt]=i;*/86         for(int i=1;i<=n;i++)87             if(!chudu[col[i]]) ans[++cnt]=i;88         sort(ans+1,ans+cnt+1);89         if(cnt) 90         {91             for(int i=1;i<cnt;i++) printf("%d ",ans[i]);92             printf("%d\n",ans[cnt]);93         }94         else printf("\n");95     }96     return 0;97 }

 

POJ——T2553 The Bottom of a Graph


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