POJ2553——The Bottom of a Graph

The Bottom of a Graph

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 8902 Accepted: 3688

Description

We will use the following (standard) definitions from graph theory. LetV be a nonempty and finite set, its elements being called vertices (or nodes). LetE 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 lengthn 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 vertexv₁ to vertex v_n+1 in G and we say thatv_n+1 is 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 nodew in G that is reachable from v, v is also reachable fromw. 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 graphG. Each test case starts with an integer number v, denoting the number of vertices ofG=(V,E), where the vertices will be identified by the integer numbers in the setV={1,...,v}. You may assume that 1<=v<=5000. That is followed by a non-negative integere 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 31 3 2 3 3 12 11 20

Sample Output

1 32

Source

Ulm Local 2003


看图就能理解，其实就是找缩点后，出度为0的点，所以就是tarjan+缩点了

#include<map>#include<set>#include<list>#include<stack>#include<queue>#include<vector>#include<cmath>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int N  = 100010;const int M  = 200010;const int inf = 0x3f3f3f3f;int DFN[N];int low[N];int block[N];int Stack[N];int out[N];bool instack[N];int head[N];int ans[N];int tot, sccnum, index, top, n, m;struct node{    int next;    int to;}edge[M];void addedge(int from, int to){    edge[tot].to = to;    edge[tot].next = head[from];    head[from] = tot++;}void init(){    memset( instack, 0, sizeof(instack) );    memset( DFN, 0, sizeof(DFN) );    memset( low, 0, sizeof(low) );    memset( out, 0, sizeof(out) );    sccnum = index = top = 0;}void tarjan(int u){    DFN[u] = low[u] = ++index;    Stack[top++] = u;    instack[u] = 1;    for (int i = head[u]; i != -1; i = edge[i].next)    {        int v = edge[i].to;        if (DFN[v] == 0)        {            tarjan(v);            if (low[u] > low[v])            {                low[u] = low[v];            }        }        else if (instack[v])        {            if (low[u] > DFN[v])            {                low[u] = DFN[v];            }        }    }    if (DFN[u] == low[u])    {        sccnum++;        do        {            top--;            block[Stack[top]] = sccnum;            instack[Stack[top]] = 0;        }while (Stack[top] != u);    }}void solve(){    init();    for (int i = 1; i <= n; i++)    {        if (DFN[i] == 0)        {            tarjan(i);        }    }    for (int i = 1; i <= n; i++)    {        for (int j = head[i]; j != -1; j = edge[j].next)        {            if (block[i] != block[edge[j].to])            {                out[block[i]]++;            }        }    }    int cnt = 0;    for (int i = 1; i <= n; i++)    {        if (out[block[i]] == 0)        {            ans[cnt++] = i;        }    }    sort(ans, ans + cnt);    printf("%d", ans[0]);    for (int i = 1; i < cnt; i++)    {        printf(" %d", ans[i]);    }    printf("\n");}int main(){    while (~scanf("%d", &n), n)    {        scanf("%d", &m);        memset( head, -1, sizeof(head) );        tot = 0;        int u, v;        for (int i = 0; i < m; i++)        {            scanf("%d%d", &u, &v);            addedge(u, v);        }        solve();    }    return 0;}