One-way connectivity of graphs

POJ 2762

Test instructions: Let me give you a diagram asking if you have a path from X to Y or from Y to X for any two points x, y?

Analysis: Is this obviously a one-way connectivity diagram? We can indent the strongly connected component, and the graph after the contraction point must be a direction-free graph; Now, the question is equivalent to giving you a undirected graph, asking if it is a one-way connected graph. In this way, we can arrange the topological ordering of the undirected undirected graph, and if there is an edge between any adjacent two points in the resulting topological sequence, that is, topo[i]--topo[i+1], the undirected graph is a unidirectional connected graph.

　　

1#include <cstdio>2#include <cstring>3#include <queue>4#include <algorithm>5 #define_CLR (x, y) memset (x, y, sizeof (x))6 #defineINF 0x3f3f3f3f7 #defineN 10108 using namespacestd;9 Ten structNode One { A     intto, next; -}edge[n*6]; - inthead[n], tot; the intDfn[n], low[n]; - intBleg[n], sta[n]; - BOOLInstack[n]; - intcnt, top, ght, N; +  - voidInit () + { Acnt=tot=ght=top=0; at_CLR (Head,-1); -_CLR (DFN,0); -_CLR (Instack,0); - } -  - voidAdd_edge (intAintb) in { -Edge[tot].to =b; toEdge[tot].next =Head[a]; +Head[a] = tot++; - } the  * voidDfsintu) $ {Panax Notoginsengdfn[u]=low[u]=++CNT; -sta[top++] =u; theInstack[u] =true; +      for(intI=head[u]; i!=-1; I=edge[i].next) A     { the         intv =edge[i].to; +         if(!Dfn[v]) -         { $ Dfs (v); $Low[u] =min (Low[u], low[v]); -         } -         Else if(Instack[v]) low[u] =min (Low[u], dfn[v]); the     } -     if(low[u]==Dfn[u])Wuyi     { theght++; -     //printf ("Num:%d\n", ght); Wu         intv; -          Do About         { $v = sta[--top]; -       //printf ("%d", v); -INSTACK[V] =false; -BLEG[V] =ght; A} while(u!=v); +     //puts (""); the     } - } $  the intInde[n], mat[n][n]; the voidTarjan () the { the      for(intI=1; i<=n; i++) -         if(!Dfn[i]) DFS (i); in  the_CLR (Inde,0); the_CLR (Mat,0); About     //to establish a new direction-free graph and to find the degree of each point the      for(intu=1; u<=n; u++) the      for(intI=head[u]; i!=-1; I=edge[i].next) the     { +         intv =edge[i].to; -         if(Bleg[v]! =Bleg[u]) the         {Bayiinde[bleg[v]]++; theMAT[BLEG[U]][BLEG[V]] =1; the         } -     } - } the  the intTopo[n], k=0; the voidTopsort () the { -queue<int>Q; the      for(intI=1; i<=ght; i++) the         if(inde[i]==0) Q.push (i); the   //printf ("Topsort: \ n");94k=0; the      while(!q.empty ()) the     { the         intU =Q.front (); Q.pop ();98     //printf ("%d", u); AboutTOPO[++K] =u; -          for(intI=1; i<=ght; i++)101         {102             if(!mat[u][i])Continue;103inde[i]--;104             if(inde[i]==0) Q.push (i); the         }106     }107    //puts ("");108 }109  the voidSolved ()111 { the     BOOLTag =true;113   //printf ("k=%d\n", k); the      for(intI=1; i<k; i++) the         if(!mat[topo[i]][topo[i+1]])//If there are two adjacent points between the topological sequences, the non-unidirectional connected graphs the         {117Tag =false;118              Break;119         } -Puts (tag?)"Yes":"No");121 }122 123 intMain ()124 { the     intT, M, A, b;126scanf"%d", &T);127      while(t--) -     {129 Init (); thescanf"%d%d", &n, &m);131          while(m--) the         {133scanf"%d%d", &a, &b);134 Add_edge (A, b);135         }136 Tarjan ();137 Topsort ();138 Solved ();139     } $     return 0;141}

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One-way connectivity of graphs