Hiho No. 215 Week Circle Detect (Topological sort | DFS)

1. For determining if there is a ring to the graph

Topology Ordering:

Topology Sorting principle:

1. From the Dag (directed acyclic graph), select a vertex without a precursor (that is, 0) and output.
2. Delete the vertex and all the forward edges that start with it.
3. Repeat 1 and 2 until the current DAG is empty or no precursor vertices exist in the current diagram, the latter case indicates that there must be a ring in the directed graph.

Time complexity: O (n+e), number of vertices and edges.

1 //topological sort to determine whether a graph has a ring2#include <queue>3#include <cstdio>4#include <cstring>5#include <algorithm>6 using namespacestd;7 8 Const intn=1e5+Ten;9 int inch[N];TenQueue <int>Q; One intt,n,m,cnt; AVector <int>E[n]; -  - voidToposort () { the      for(intI=1; i<=n;i++){ -         if(inch[i]==0) Q.push (i); -     } -      while(!Q.empty ()) { +cnt++; -         intu=Q.front (); + Q.pop (); A          for(intI=0; I<e[u].size (); i++){ at             intv=E[u][i]; -             inch[v]--; -             if(inch[v]==0) Q.push (v); -         } -     } - } in  - intMain () { toscanf"%d",&t); +      while(t--){ -Cnt=0; theMemsetinch,0,sizeof(inch)); *          for(intI=1; i<n;i++) e[i].clear (); $scanf"%d%d",&n,&m);Panax Notoginseng          for(intI=1; i<=m;i++){ -             intx, y; thescanf"%d%d",&x,&y); + e[x].push_back (y); A             inch[y]++; the         } + Toposort (); -         if(cnt==n) printf ("no\n"); $         Elseprintf"yes\n"); $     } -     return 0; -}

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Dfs:

Point staining during DFS:

1. The points that have not been accessed by DFS are white, and the initial is a little white.
2. If the point u has been accessed by DFS, but the child node of U is not fully accessible, then you are dyed grey
3. If both the point U and the child nodes of u have been visited, you will be dyed black when you go back to the parent node of U.

If in the process of Dfs we reach a gray node along a forward edge, there is a ring in the diagram, and if the gray node has never been reached, there is no ring.

Time complexity: O (n+e), number of vertices and edges.

1 //DFS Determines if the graph has a ring2#include <vector>3#include <cstdio>4#include <cstring>5#include <iostream>6#include <algorithm>7 using namespacestd;8 9 Const intn=1e5+Ten;TenVector <int>E[n]; One intF,vis[n]; A  - voidDfsintu) { -     if(f)return ; thevis[u]=-1; -      for(intI=0; I<e[u].size (); i++){ -         intv=E[u][i]; -         if(vis[v]==0) Dfs (v); +         Else if(vis[v]==-1) {f=1;return ;} -     } +vis[u]=1; A } at  - intMain () { -     intt,n,m; -scanf"%d",&t); -      while(t--){ -f=0; inmemset (Vis,0,sizeof(Vis)); -          for(intI=1; i<n;i++){ to e[i].clear (); +         } -scanf"%d%d",&n,&m); the         intb; *          for(intI=1; i<=m;i++){ $scanf"%d%d",&a,&b);Panax Notoginseng E[a].push_back (b); -         } the          for(intI=1; i<=n;i++){ +             if(f) Break; A             if(vis[i]==0) Dfs (i); the         } +         if(f==1) printf ("yes\n"); -         Elseprintf"no\n"); $     } $     return 0; -}

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Note: Note the emptying of the vector's storage edge.

2. Determine if the graph has a ring

Suppose there are n connected sub-graphs of undirected graphs. If undirected graphs have no loops, each connected sub-graph is a tree, that is, the number of vertices of the connected sub-graph (vi) = number of sides (EI) +1.

Last obtained: V=e+n

Hiho No. 215 Week Circle Detect (Topological sort | DFS)