POJ 2553 The Bottom of a Graph(强连通分量)

思路： 强连通分量题的老套路:求分量,缩点图,输出DAG的相关信息.

该题的答案就是那些DAG中出度为0的点 所代表的分量中包含的所有点.

#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
#include <stack>
using namespace std;
#define maxn 10000+100
#define LL long long
int cas=1,T;
vector<int>G[maxn];
int pre[maxn];
int lowlink[maxn];
int sccno[maxn];
int num[maxn];            //在i编号scc中有多少个点
int dfs_clock,scc_cnt;
int n,m;
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 v = G[u][i];
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;
num[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 in[maxn];
int out[maxn];
int main()
{
//freopen("in","r",stdin);
while (scanf("%d",&n)==1 && n)
{
scanf("%d",&m);
for (int i = 0;i<=n;i++)
G[i].clear();
memset(out,0,sizeof(out));
memset(num,0,sizeof(num));
for (int i = 1;i<=m;i++)
{
int u,v;
scanf("%d%d",&u,&v);
G[u].push_back(v);
}
find_scc(n);
for (int i = 1;i<=scc_cnt;i++)
{
out[i]=true;
}
for (int u = 1;u<=n;u++)
for (int i =0;i<G[u].size();i++)
{
int v = G[u][i];
if (sccno[u] != sccno[v])
out[sccno[u]]=false;
}
int b=0;

int flag = 1;
for (int i = 1;i<=n;i++)
{
if (out[sccno[i]])
{
if (flag)
printf("%d",i),flag=0;
else
printf(" %d",i);
}
}
puts("");
}
return 0;
}

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 that1<=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