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(500010);
const int N(5010);
int n,m,u,v;
int sumedge,head[N];
struct Edge
{
    int to,next;
    Edge(int to=0,int next=0) :
        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=0;
    sumcol=sumedge=0;
    memset(ans,0,sizeof(ans));
    memset(low,0,sizeof(low));
    memset(dfn,0,sizeof(dfn));
    memset(col,0,sizeof(col)); 
    memset(head,0,sizeof(head)); 
    memset(chudu,0,sizeof(chudu));
    memset(Stack,0,sizeof(Stack)); 
    memset(instack,0,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=1;i<=n;i++)
            if(!dfn[i]) DFS(i);
        for(u=1;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=1;i<=n;i++)
            if(!chudu[col[i]]) ans[++cnt]=i;
        sort(ans+1,ans+cnt+1);
        if(cnt) 
        {
            for(int i=1;i<cnt;i++) printf("%d ",ans[i]);
            printf("%d
",ans[cnt]);
        }
        else printf("
");
    }
    return 0;
}