POJ 2553 The Bottom of a Graph

Posted 2020-08-25 り挽歌、花开花落的流年

 

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 10687		Accepted: 4403

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. ThenG=(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+1is 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

Source

Ulm Local 2003

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<vector>
 5 #include<algorithm> 
 6 #include<stack>
 7 #define maxm 500010
 8 #define maxn 5010
 9 using namespace std;
10 int T,n,m;
11 struct edge{
12     int u,v,next;
13 }e[maxm],ee[maxm];
14 vector<int> kp[maxn],ans;
15 int head[maxn],js,headd[maxn],jss;
16 bool exist[maxn];
17 int visx,cur;// cur--缩出的点的数量 
18 int dfn[maxn],low[maxn],belong[maxn],chudu[maxn];
19 stack<int>st;
20 void init(){
21     memset(chudu,0,sizeof(chudu));
22     memset(e,0,sizeof(e));memset(ee,0,sizeof(ee));
23     memset(head,0,sizeof(head));memset(headd,0,sizeof(headd));
24     jss=js=visx=cur=0;
25     memset(exist,false,sizeof(exist));
26     while(!st.empty())st.pop();
27     memset(dfn,0,sizeof(dfn));memset(low,0,sizeof(low));
28     memset(belong,0,sizeof(belong));
29     ans.resize(0);
30     for(int i=0;i<maxn;i++)kp[i].resize(0);
31 }
32 void add_edge1(int u,int v){
33     e[++js].u=u;e[js].v=v;
34     e[js].next=head[u];head[u]=js;
35 }
36 void tarjan(int u){
37     dfn[u]=low[u]=++visx;
38     exist[u]=true;
39     st.push(u);
40     for(int i=head[u];i;i=e[i].next){
41         int v=e[i].v;
42         if(dfn[v]==0){
43             tarjan(v);
44             low[u]=min(low[u],low[v]);
45         }
46         else if(exist[v]&&low[u]>dfn[v]) low[u]=dfn[v];
47     }
48     int j; 
49     if(low[u]==dfn[u]){
50         ++cur;
51         do{
52             j=st.top();st.pop();exist[j]=false;
53             kp[cur].push_back(j);
54             belong[j]=cur;
55         }while(j!=u);
56     }
57 }
58 void add_edge2(int u,int v){
59     ee[++jss].u=u;ee[jss].v=v;
60     ee[jss].next=headd[u];headd[u]=jss;
61 }
62 void apki(int x){
63     for(int i=0;i<kp[x].size();i++){
64         int k=kp[x][i];
65         ans.push_back(k);
66     }
67 }
68 int main()
69 {
70     while(scanf("%d%d",&n,&m)==2){
71         init();
72         int u,v;
73         for(int i=0;i<m;i++){
74             scanf("%d%d",&u,&v);add_edge1(u,v);
75         }
76         for(int i=1;i<=n;i++){// 求强连通分量
77             if(dfn[i]==0) tarjan(i);
78         }
79         for(int i=1;i<=m;i++){
80             int u=e[i].u,v=e[i].v;
81             if(belong[u]!=belong[v]){
82                 add_edge2(belong[u],belong[v]);
83                 chudu[belong[u]]++;
84             }
85         }
86         for(int i=1;i<=cur;i++){
87             if(chudu[i]==0)
88               apki(i);
89         }
90         sort(ans.begin(),ans.end());
91         for(int i=0;i<ans.size();i++) printf("%d ",ans[i]);
92         printf("\n");
93     }
94     return 0;
95 }

 思路：和POJ2762差不很多