The Bottom of a Graph poj 2553 缩点+Tarjan

The Bottom of a Graph

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 8386		Accepted: 3460

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

这个题目是很棒的，主要的部分在于. 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. 题目是找sink,通过题意可以知道就是找强联通变量，找出出度为0的点，然后输出来，直接套Tarjan的模板，稍微改一下就行，比较水的

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define M 50010
#define N 20010
struct node
{
int v,next;

} unit[M];
int first
,stack
,DFN
,Low
,B
;
int instack
;
int in
,out
;
int n,m,c,sc,top,num;

void init()
{
c=0;
sc=top=num=0;
memset(first,-1,sizeof(first));
memset(B,0,sizeof(B));
memset(DFN,0,sizeof(DFN));
memset(out,0,sizeof(out));
memset(in,0,sizeof(in));
}
int min(int a, int b)
{
return a < b ? a : b;
}

int max(int a, int b)
{
return a > b ? a : b;
}
void add(int a,int b)
{
unit[num].v=b;
unit[num].next=first[a];
first[a]=num++;
}

void Tarjan(int v)
{

int min,t,e,j;
DFN[v]=Low[v]=++num;
instack[v]=1;
stack[top++]=v;
for(e=first[v]; e!=-1; e=unit[e].next)
{
j=unit[e].v;
if(!DFN[j])
{
Tarjan(j);
if(Low[v]>Low[j])
{
Low[v]=Low[j];
}

}
else if(instack[j]&&DFN[j]<Low[v])
{
Low[v]=DFN[j];
}
}
if(DFN[v]==Low[v])
{
while(v!=stack[top])
{

B[stack[top-1]]=sc;

instack[stack[top-1]]=0;

top--;

}

sc++;
}

}
void solve()
{

int i;
for(i=1; i<=n; i++)
{
if(!DFN[i])
{
Tarjan(i);
}
}
}

int main()
{
while(scanf("%d",&n)!=EOF)
{

if(n==0)
break;
init();
scanf("%d",&m);
int i,j,a,b;
for(i=0;i<m;i++)
{
scanf("%d%d",&a,&b);
add(a,b);
}
//printf("fdhskfjhdskfjfdsfsd");
for(int i=1;i<=n;i++)
if(!DFN[i])
Tarjan(i);
for(int i=1; i<=n; i++)
{
for(int end=first[i]; end!=-1; end=unit[end].next)
{
if(B[i]!=B[unit[end].v])
{

//in[B[unit[end].v]]++;
// printf("aaaaaaaaa");
out[B[i]]++;

}

}

}

int maxin=0,maxout=0;
// printf("ccc  %d\n",sc);
/*    for(i=0;i<sc;i++)
{
printf("%d  %d\n",in[i],out[i]);
}*/
int flag=1;
//printf("fdsjl");
for(i=1;i<=n;i++)
{
if(!out[B[i]])
{
if(flag)
{
printf("%d",i);
flag=0;
}
else
printf(" %d",i);
}

}

printf("\n");
}

return 0;
}