POJ——T2553 The Bottom of a Graph

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

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

 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 Edge
12 {
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 }