The Bottom of a Graph
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
题意：求出强连通分量，判断每个分量是否有出边。按序输出没出边的分量里的点。
代码：
#include <stdio.h>
#include <stdlib.h>
#define Min(a, b) a < b ? a : b
#define maxn 5001
struct T
{
int v, next;
}fn[maxn * maxn];
int g[maxn], visit[maxn], low[maxn], dfn[maxn], stack[maxn], f[maxn], flag[maxn];
int top, id;
void set(int n)
{
for (int i = 1; i <= n; i++)
{
g[i] = -1, visit[i] = 0, f[i] = 0, flag[i] = 0;
}
}
int dfs(int u, int t)
{
int i;
visit[u] = 1;
stack[top++] = u;
dfn[u] = low[u] = t;
for (i = g[u]; i != -1; i = fn[i].next)
{
if (!visit[fn[i].v])
{
t = dfs(fn[i].v, t + 1);//返回遍历完u的子树的时间
low[u] = Min(low[u], low[fn[i].v]);//如果有孩子因为后向边被缩小了low值，则它也相应缩小low值，整个分量的low都取分量中low最小的值。
}
else
{
if (!f[fn[i].v])//如果f[v]还是0，表示u->v是后向边，v还没出过栈过
{
low[u] = Min(low[u], low[fn[i].v]);
}
}
}
if (dfn[u] == low[u])
{
id++;
do
{
f[stack[--top]] = id;
}while (top > 0 && stack[top] != u);
}
return t;
}
void tarjan(int n)
{
int t = 0;
id = 0;
int i;
for (i = 1; i <= n; i++)
{
if (!visit[i])
{
//如果i点能与其它点x构成一个分量，则一定能搜到x，否则，它自己便是一个分量，所以这里的搜索不会重复。
t = dfs(i, t + 1);
}
}
}
int main()
{
int n, m, u, v, th, k, i, j;
while (scanf("%d", &n), n)
{
scanf("%d", &m);
th = 0;
set(n);
while (m--)
{
scanf("%d%d", &u, &v);
fn[th].v = v, fn[th].next = g[u], g[u] = th++;
}
tarjan(n);
for (i = 1; i <= n; i++)
{
for (j = g[i]; j != -1; j = fn[j].next)
{
if (f[fn[j].v] != f[i])//两个点的所在分量不同，表示i有出边
{
//hash[i] = 1;
flag[f[i]] = 1;
break;
}
}
}
for (i = 1; i <= n; i++)
{
if (!flag[f[i]])
{
if (i != n)
{
printf("%d ", i);
}
else
{
printf("%d", i);
}
}
}
printf("\n");
}
system("pause");
return 0;
}
/*
6 8
1 3
1 2
3 5
5 6
2 4
4 6
3 4
4 1
4 5
1 2
2 3
3 1
1 3
3 4
*/