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 vv 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

Source

 
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 ;
}

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