poj2553The Bottom of a Graph(强连通+缩点)

The Bottom of a Graph

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 7699 Accepted: 3161

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. Then G=(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+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 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 31 3 2 3 3 12 11 20

Sample Output

1 32

Source

Ulm Local 2003

题目大意：给定一个有向图，求一个点集，要求这个点集里的所有点能到达的点，也都能到达这个点。

题目分析：就是求强连通嘛，不过并不是求所有的强连通。因为强连通之间也可能有边相连。将强连通缩点后就将一张有向图转化成DAG，只能将其中出度为0的强连通分量输出，因为只有这些强连通分量中的点能够相互到达。其他的强连通分量可以到达这个强连通分量，但是回不去了，所以不符合。

详情请见代码：

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N = 5005;const int M = 1000005;struct edge{    int to,next;}g[M];int head[N];int scc[N];int stack1[N];int stack2[N];int out[N];int vis[N];int ans[N];int n,m;void init(){    for(int i = 1;i <= n;i ++)    {        head[i] = -1;        scc[i] = out[i] = vis[i] = 0;    }}void dfs(int cur,int &sig,int &ret){    vis[cur] = ++ sig;    stack1[++stack1[0]] = cur;    stack2[++stack2[0]] = cur;    for(int i = head[cur];i != -1;i = g[i].next)    {        if(vis[g[i].to] == 0)            dfs(g[i].to,sig,ret);        else            if(scc[g[i].to] == 0)            {                while(vis[stack2[stack2[0]]] > vis[g[i].to])                    stack2[0] --;            }    }    if(stack2[stack2[0]] == cur)    {        ++ret;        stack2[0] --;        do        {            scc[stack1[stack1[0]]] = ret;        }while(stack1[stack1[0] --] != cur);    }}int Gabow(){    int i,sig,ret;    stack1[0] = stack2[0] = sig = ret = 0;    for(i = 1;i <= n;i ++)        if(!vis[i])            dfs(i,sig,ret);    return ret;}void solve(){    int i,j,num;    num = Gabow();    if(num == 1)    {        for(i = 1;i < n;i ++)            printf("%d ",i);        printf("%d\n",i);        return;    }    for(i = 1;i <= n;i ++)    {        for(j = head[i];j != -1;j = g[j].next)        {            if(scc[i] != scc[g[j].to])                out[scc[i]] ++;        }    }    int ansnum = 0;    for(i = 1;i <= num;i ++)    {        if(out[i] == 0)        {            for(j = 1;j <= n;j ++)                if(scc[j] == i)                    ans[ansnum ++] = j;        }    }    sort(ans,ans + ansnum);    for(i = 0;i < ansnum - 1;i ++)        printf("%d ",ans[i]);    printf("%d\n",ans[i]);}int nextint(){    char c;    int ret;    while((c = getchar()) > '9' || c < '0')        ;    ret = c - '0';    while((c = getchar()) >= '0' && c <= '9')        ret = ret * 10 + c - '0';    return ret;}int main(){    int i,j,a,b;    while(n = nextint(),n)    {        m = nextint();        init();        for(i = 1;i <= m;i ++)        {            a = nextint();            b = nextint();            g[i].to = b;            g[i].next = head[a];            head[a] = i;        }        solve();    }    return 0;}//996K110MS