Poj 2533 The Bottom of a Graph【强连通Tarjan】

The Bottom of a Graph

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 10099		Accepted: 4175

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：给你一堆点，一堆边，让你找到缩点之后出度为0的节点， 然后将节点编号从小到大排序输出。

思路：Tarjan，缩点染色，判断出度为0的强连通分量，将整个集合排序，输出即可。

AC代码：

#include<stdio.h>
#include<string.h>
#include<vector>
#include<algorithm>
using namespace std;
#define maxn 1000000
vector<int >mp[maxn];
int ans[maxn];
int degree[maxn];
int color[maxn];
int stack[maxn];
int vis[maxn];
int dfn[maxn];
int low[maxn];
int n,m,tt,cnt,sig;
void init()
{
memset(degree,0,sizeof(degree));
memset(color,0,sizeof(color));
memset(stack,0,sizeof(stack));
memset(low,0,sizeof(low));
memset(dfn,0,sizeof(dfn));
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++)mp[i].clear();
}
void Tarjan(int u)
{
vis[u]=1;
low[u]=dfn[u]=cnt++;
stack[++tt]=u;
for(int i=0;i<mp[u].size();i++)
{
int v=mp[u][i];
if(vis[v]==0)Tarjan(v);
if(vis[v]==1)low[u]=min(low[u],low[v]);
}
if(dfn[u]==low[u])
{
sig++;
do
{
color[stack[tt]]=sig;
vis[stack[tt]]=-1;
}
while(stack[tt--]!=u);
}
}
void Slove()
{
tt=-1;cnt=1;sig=0;
for(int i=1;i<=n;i++)
{
if(vis[i]==0)
{
Tarjan(i);
}
}
for(int i=1;i<=n;i++)
{
for(int j=0;j<mp[i].size();j++)
{
int v=mp[i][j];
if(color[i]!=color[v])
{
degree[color[i]]++;
}
}
}
int tot=0;
for(int i=1;i<=sig;i++)
{
if(degree[i]>0)continue;
for(int j=1;j<=n;j++)
{
if(color[j]==i)
{
ans[tot++]=j;
}
}
}
sort(ans,ans+tot);
for(int i=0;i<tot;i++)
{
if(i==0)printf("%d",ans[i]);
else printf(" %d",ans[i]);
}
printf("\n");
}
int main()
{
while(~scanf("%d",&n))
{
if(n==0)break;
scanf("%d",&m);
init();
for(int i=0;i<m;i++)
{
int x,y;
scanf("%d%d",&x,&y);
mp[x].push_back(y);
}
Slove();
}
}