poj 2553 The Bottom of a Graph

[align=center]The Bottom of a Graph[/align]

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 10912		Accepted: 4497

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：如果v点能够到的点，反过来能够到达v点，则称这个点为sink点，输出所有的sink点

解题思路：求连通分量，然后出度为0的连通分量里面的点就是sink点

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>
using namespace std;
#define MAX 100005
#define MAXN 500005
struct Edge
{
int u;
int v;
int next;
} edge[MAXN];
int DFN[MAX], low[MAX], in[MAX], out[MAX];
int flag[MAX], step[MAX], head[MAX];
int sum[MAX];
stack<int>S;
int res, tot, M, ans, ans1, N;
void Init()
{
memset(DFN, 0, sizeof(DFN));
memset(low, 0, sizeof(low));
memset(in, 0, sizeof(in));
memset(out, 0, sizeof(out));
memset(flag, 0, sizeof(flag));
memset(head, -1, sizeof(head));
memset(sum, 0, sizeof(sum));
memset(edge, 0, sizeof(edge));
memset(step, 0, sizeof(step));
if(!S.empty())
S.pop();
tot = 0, res = 0;
}
void addedge(int u, int v, int k)
{
edge[k].u = u, edge[k].v = v, edge[k].next = head[u], head[u] = k;
}
void tarjan(int u)
{
DFN[u] = low[u] = ++tot;
flag[u] = 1;
S.push(u);
for(int j = head[u]; j != -1; j = edge[j].next) ///更新low值
{
int v = edge[j].v;
if(!DFN[v])
{
tarjan(v);
low[u] = min(low[u],low[v]);
}
else if(flag[v])
{
low[u] = min(low[u],low[v]);
}
}

int v;
if(DFN[u]==low[u]) ///缩点
{
do
{
v = S.top();
S.pop();
step[v] = res;
sum[res]++;
flag[v] = 0;
}
while(u!=v);
res++;
}
}
void solve()
{
for(int i=0; i<N; i++)
for(int j=head[i]; j!=-1; j=edge[j].next) ///统计缩点后的各点的出度
{
if(step[i]!=step[edge[j].v])
out[step[i]]++;
}
int f, yy=0;
for(int i = 0; i < N; i++)
{
if(!out[step[i]])
flag[i] = 1;
}
for(int i = 0; i < N; i++)
{
if(flag[i])
printf("%d ",i + 1);
}
printf("\n");
}
int main()
{
while(~scanf("%d",&N)&&N)
{
scanf("%d",&M);
Init();
for(int i = 0; i < M; i++)
{
int a, b;
scanf("%d%d",&a,&b);
addedge(a-1,b-1,i); ///建图
}
for(int i = 0; i < N; i++)
{
if(!DFN[i])
{
tarjan(i);
}
}
solve();
}
return 0;
}