POJ 2553——The Bottom of a Graph（强连通分量）

The Bottom of a Graph
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 10222 Accepted: 4229
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,…,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+1 is 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 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,…,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

Ulm Local 2003

题目大意：给出一个有向图，求出图中哪些点满足——它能到达的点都能到达它——并输出各点。

题解：本题是一个求强连通分量的题目。显然在同一强连通分量中的点，如果满足性质则都满足，反之亦然。又因为如果某个强连通分量不满足性质，则必然存在一条由本连通分量指向其他分量的边，故所求的满足条件的强连通分量出度为0。求出所有出度为0的强连通分量即可。
求强连通分量可用Tarjan算法——详见白书。

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <stack>#include <vector>#include <cassert>using namespace std;const int maxn = 5000+50;const int maxm = 1e6+10;struct Edge{    int u;    int v;    int w;}edges[maxm];vector<int> G[maxn];int pre[maxn];int sccno[maxn];            //每个点所在的强连通分量的编号 int lowlink[maxn];int mark[maxn];int is_ans[maxn];int in0[maxn];int out0[maxn];int n,m,ans,a_from;int dfs_clock,scc_cnt;stack<int> S; void dfs(int u){    pre[u]=lowlink[u]=++dfs_clock;    S.push(u);    for(int i=0;i<G[u].size();i++){        int e = G[u][i];        int v = edges[e].v;        if(!pre[v]){            dfs(v);            lowlink[u] = min(lowlink[u],lowlink[v]);        }        else if(!sccno[v]){            lowlink[u] = min(lowlink[u],pre[v]);        }    }    if(lowlink[u] == pre[u]){        scc_cnt++;        for(;;){            int x = S.top();S.pop();            sccno[x] = scc_cnt;            if(x == u) break;        }    }}void find_scc(int n){    dfs_clock = scc_cnt = 0;    memset(sccno,0,sizeof(sccno));    memset(pre,0,sizeof(pre));    for(int i=1;i<=n;i++)      if(!pre[i]) dfs(i);}int main(){    freopen("A.in","r",stdin);    freopen("A.out","w",stdout);    while(~scanf("%d%d",&n,&m) && n){        ans = 0;dfs_clock=0;        memset(pre,0,sizeof(pre));        memset(sccno,0,sizeof(sccno));        memset(lowlink,0,sizeof(lowlink));        memset(is_ans,0,sizeof(is_ans));            for(int i=1;i<=n;i++) G[i].clear();        for(int i=1;i<=m;i++){            scanf("%d%d",&edges[i].u,&edges[i].v);            G[edges[i].u].push_back(i);        }        find_scc(n);        for(int i=1;i<=scc_cnt;i++)  in0[i] = out0[i] = 1;        for(int i=1;i<=n;i++)          for(int j=0;j<G[i].size();j++){            int e =  G[i][j];            int v = edges[e].v;            if(sccno[i]!=sccno[v]) out0[sccno[i]] = in0[sccno[v]]= 0;          }        for(int i=1;i<=n;i++)   if(out0[sccno[i]]) printf("%d ",i);        printf("\n");    }    return 0;}