Poj 2553 taarjan contraction point + degree

The bottom of a graph

Time limit:3000 Ms		Memory limit:65536 K
Total submissions:8904		Accepted:3689

Description

We will use the following (standard) Definitions from graph theory. Let VBe a nonempty and Finite Set, its elements being called vertices (or nodes). Let EBe a subset of the Cartesian Product V × V, Its elements being called edges. Then G = (V, E)Is called a directed graph.
Let NBe a positive integer, And let P = (E1,..., en)Be a sequence of Length NOf Edges Ei, ESuch that Ei = (Vi, vi + 1)For a sequence of vertices (V1,..., vn + 1). Then PIs called a path from vertex V1To Vertex Vn + 1In GAnd we say that Vn + 1Is reachable from V1, Writing (V1 → VN + 1).
Here are some new definitions. A node VIn a graph G = (V, E)Is called a sink, if for every node WIn GThat is reachable from V, VIs also reachable from W. The bottom of a graph is the subset of all nodes that are sinks, I. e ., Bottom (G) = {v ε v | percentile wε v :( v → W) percentile (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 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 EAnd, thereafter, EPairs of vertex identifiers V1, W1,..., VE, weWith the meaning that (Vi, WI) ε 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 31 3 2 3 3 12 11 20

Sample output

1 32


Question meaning:
For a directed graph with n knots and m edges, if any other vertex that a vertex can reach the vertex or a vertex cannot reach another vertex, the vertex is sink, from small to output sink.


Ideas:
This question is difficult in translation, especially the four I took two tests to the people T-T, understand it is very water, first use Tarjan shrink point block, then we can mark the points of the blocks without degree, and finally sort the output.


Code:

 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 500511 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 }

 

Poj 2553 taarjan contraction point + degree