POJ 2553 taarjan缩点+度数

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,...,e_n) be a sequence of length n of edges e_i∈E such that e_i=(v_i,v_i+1) for a sequence of vertices (v₁,...,v_n+1). Then p is called a path from vertex v₁ to vertex v_n+1 in G and we say that v_n+1 is reachable from v₁, writing (v₁→v_n+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,...,v_e,w_e with the meaning that (v_i,w_i)∈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

题目意思：给一个n个结点，m条边的有向图，若一个点所能到达的其他点都能到达该点或一个点不能到达其他点，那么该点为sink，从小到达输出sink。

思路：这个题难在翻译上，特别是我这四级考了两次才过的人T-T，看懂了就很水了，先用tarjan缩点分块，然后求出没有出度的块的点标记下来，最后排序输出就行了。

代码：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <iostream>
 5 #include <vector>
 6 #include <queue>
 7 #include <stack>
 8 using namespace std;
 9
10 #define N 5005
11 int n, m;
12 vector<int>ve[N];
13 int low[N], dfn[N], dfn_clock;
14 int in[N], out[N], instack[N], a[N], b[N];
15 int cnt;
16 stack<int>st;
17
18 void init(){
19     int i, j;
20     while(!st.empty()) st.pop();
21
22     for(i=0;i<=n;i++){
23         ve[i].clear();instack[i]=1;
24     }
25     memset(dfn,0,sizeof(dfn));
26     memset(in,0,sizeof(in));
27     memset(out,0,sizeof(out));
28 }
29
30 void tarjan(int u){
31     int i, j, v;
32     dfn[u]=low[u]=dfn_clock++;
33     st.push(u);
34     for(i=0;i<ve[u].size();i++){
35         v=ve[u][i];
36         if(!dfn[v]){
37             tarjan(v);
38             low[u]=min(low[u],low[v]);
39         }
40         else if(instack[v]){
41             low[u]=min(low[u],dfn[v]);
42         }
43     }
44     if(low[u]==dfn[u]){
45         cnt++;
46         while(1){
47             v=st.top();st.pop();
48             a[v]=cnt;instack[v]=0;
49             if(v==u) break;
50         }
51     }
52 }
53
54 main()
55 {
56     int i, j, k;
57     int x, y;
58     while(scanf("%d",&n)==1&&n){
59         scanf("%d",&m);
60         init();
61         while(m--){
62             scanf("%d %d",&x,&y);
63             ve[x].push_back(y);
64         }
65         dfn_clock=1;cnt=0;
66         for(i=1;i<=n;i++){
67             if(!dfn[i])
68             tarjan(i);
69         }
70         for(i=1;i<=n;i++){
71             for(j=0;j<ve[i].size();j++){
72                 x=ve[i][j];
73                 if(a[x]!=a[i]){
74                     out[a[i]]++;in[a[x]]++;
75                 }
76             }
77         }
78         k=0;
79         for(i=1;i<=n;i++){
80             if(!out[a[i]]){
81                 b[k++]=i;
82             }
83         }
84         sort(b,b+k);
85         printf("%d",b[0]);
86         for(i=1;i<k;i++) printf(" %d",b[i]);
87         cout<<endl;
88     }
89 }