poj2553The Bottom of a Graph(强连通+缩点)
The Bottom of a Graph
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 7699 | Accepted: 3161 |
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.
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：给定一个有向图，求一个点集，要求这个点集里的所有点能到达的点，也都能到达这个点。
题目分析：就是求强连通嘛，不过并不是求所有的强连通。因为强连通之间也可能有边相连。将强连通缩点后就将一张有向图转化成DAG，只能将其中出度为0的强连通分量输出，因为只有这些强连通分量中的点能够相互到达。其他的强连通分量可以到达这个强连通分量，但是回不去了，所以不符合。
详情请见代码：
#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 5005;
const int M = 1000005;
struct edge
{
int to,next;
}g[M];
int head[N];
int scc[N];
int stack1[N];
int stack2[N];
int out[N];
int vis[N];
int ans[N];
int n,m;
void init()
{
for(int i = 1;i <= n;i ++)
{
head[i] = -1;
scc[i] = out[i] = vis[i] = 0;
}
}
void dfs(int cur,int &sig,int &ret)
{
vis[cur] = ++ sig;
stack1[++stack1[0]] = cur;
stack2[++stack2[0]] = cur;
for(int i = head[cur];i != -1;i = g[i].next)
{
if(vis[g[i].to] == 0)
dfs(g[i].to,sig,ret);
else
if(scc[g[i].to] == 0)
{
while(vis[stack2[stack2[0]]] > vis[g[i].to])
stack2[0] --;
}
}
if(stack2[stack2[0]] == cur)
{
++ret;
stack2[0] --;
do
{
scc[stack1[stack1[0]]] = ret;
}while(stack1[stack1[0] --] != cur);
}
}
int Gabow()
{
int i,sig,ret;
stack1[0] = stack2[0] = sig = ret = 0;
for(i = 1;i <= n;i ++)
if(!vis[i])
dfs(i,sig,ret);
return ret;
}
void solve()
{
int i,j,num;
num = Gabow();
if(num == 1)
{
for(i = 1;i < n;i ++)
printf("%d ",i);
printf("%d\n",i);
return;
}
for(i = 1;i <= n;i ++)
{
for(j = head[i];j != -1;j = g[j].next)
{
if(scc[i] != scc[g[j].to])
out[scc[i]] ++;
}
}
int ansnum = 0;
for(i = 1;i <= num;i ++)
{
if(out[i] == 0)
{
for(j = 1;j <= n;j ++)
if(scc[j] == i)
ans[ansnum ++] = j;
}
}
sort(ans,ans + ansnum);
for(i = 0;i < ansnum - 1;i ++)
printf("%d ",ans[i]);
printf("%d\n",ans[i]);
}
int nextint()
{
char c;
int ret;
while((c = getchar()) > '9' || c < '0')
;
ret = c - '0';
while((c = getchar()) >= '0' && c <= '9')
ret = ret * 10 + c - '0';
return ret;
}
int main()
{
int i,j,a,b;
while(n = nextint(),n)
{
m = nextint();
init();
for(i = 1;i <= m;i ++)
{
a = nextint();
b = nextint();
g[i].to = b;
g[i].next = head[a];
head[a] = i;
}
solve();
}
return 0;
}
//996K 110MS