POJ2553————The Bottom of a Graph（tarjan算法）

The Bottom of a Graph

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 10668		Accepted: 4391

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

大意：题目理解的难点在于什么是sinks，也就是比如有u,v两个点，如果u能到v，能够推出v能到u,那么u,v就是sinks。

比如第二个样例,1能到2,但是2不能到1,所以1不是sink，但是由于2谁也到不了，所以2也是sink

那么也就是相当于求所有的初度为0的强连通分量

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
using namespace std;

const int MAXN = 5010;
int stack[MAXN];
int top=0;
bool vis[MAXN];
bool used[MAXN];
int dfn[MAXN];
int low[MAXN];
int ComNum = 0;
int indexx=0;
vector <int> G[MAXN];
vector <int> Com[MAXN];
int InCom[MAXN];
int out_dgree[MAXN];
vector <int> res;

void init(int n){
top=0;
memset(vis,0,sizeof(vis));
memset(used,0,sizeof(used));
fill(dfn,dfn+MAXN,-1);
fill(low,low+MAXN,-1);
ComNum=0;
indexx=0;
for(int i=1;i<=n;i++){
G[i].clear();
Com[i].clear();
}
memset(InCom,0,sizeof(InCom));
memset(out_dgree,0,sizeof(out_dgree));
res.clear();
}

void tarjan(int i){
dfn[i]=low[i]=++indexx;
vis[i]=true;
used[i]=true;
stack[++top]=i;
for(int k=0;k<G[i].size();k++){
int v=G[i][k];
if(dfn[v]==-1){
tarjan(v);
low[i]=min(low[i],low[v]);
}
else if(vis[v]){
low[i]=min(low[i],dfn[v]);
}
}
int j;
if(dfn[i]==low[i]){
ComNum++;
do{
j=stack[top--];
vis[j]=false;
Com[ComNum].push_back(j);
InCom[j]=ComNum;
}
while(j!=i);
}
}

int main()
{
int n,m;
while(scanf("%d",&n)){
if(n==0)
break;
scanf("%d",&m);
init(n);
while(m--){
int u,v;
scanf("%d%d",&u,&v);
G[u].push_back(v);
}
for(int i=1;i<=n;i++){
if(!used[i])
tarjan(i);
}
for(int u=1;u<=n;u++){
for(int j=0;j<G[u].size();j++){
int v=G[u][j];
if(InCom[u]!=InCom[v]){
out_dgree[InCom[u]]++;
}
}
}
for(int i=1;i<=ComNum;i++){
if(out_dgree[i]==0){
for(int j=0;j<Com[i].size();j++){
res.push_back(Com[i][j]);
}
}
}
sort(res.begin(),res.end());
for(int i=0;i<res.size();i++){
if(i==res.size()-1)
printf("%d",res[i]);
else
printf("%d ",res[i]);
}
printf("\n");
}
}