"Topological sorting problems"
Workaround:
1. Calculate the entry value for each point deg[i], this step needs to scan all points and edges, complexity O (n+m).
2. Adding a point of 0 to the queue Q, of course there may be more than 0 points, and there is no connection between them, so it is possible to add Q in any order.
3. Remove a point p from the Q. For each point not deleted and connected to P q,deg[q] = deg[q]-1; if deg[q]==0, add Q to Q.
Code: Complexity: O (v+e)
<span style= "FONT-SIZE:14PX;" > #include <bits/stdc++.h>using namespace std;const int N=5*1e5+10;const int Mod=142857;int t,n,k,m,x;int Father[n],v[n],indegree[n];vector <int >vec[n];bool Topsort () {Queue<int >q; while (!q.empty ()) Q.pop (); for (int i=1; i<=n; i++) if (indegree[i]==0) Q.push (i); int ans=0,sum=0; while (!q.empty ()) {int U=q.front (); Q.pop (); sum++; for (int i=0; i<vec[u].size (); i++) {int temp=vec[u][i]; if (--indegree[temp]==0) Q.push (temp); }} if (Sum<n) return false;//to determine the topological sort return true; int main () {int u,v,a; scanf ("%d", &t); while (t--) {scanf ("%d%d", &n,&m); memset (indegree,0,sizeof (Indegree)); for (int i=1; i<=n; i++) if (Vec[i].size ()) vec[i].clear (); while (m--) {scanf ("%d%d", &v,&u); Vec[v].push_back (U); indegree[u]++; } if (topSort ()) puts ("YES"); Else puts ("NO"); } return 0;} </span>
If the time complexity of topological sorting is O (n*e) with adjacency table, the adjacency matrix is O (n^2), N is the number of vertices, E is the number of edges, and the Floyd time Complexity is O (n^3).
Properties
1, topological sequencing in the direction of the non-circular graph to discharge the effective sequence, otherwise it can be judged that there is a ring to the graph.
2, if the input of the point in the graph, there is no point in the degree of 0, then the direction of the graph exists loop
3, if the presence of a point of 0 is greater than one, then the graph certainly does not exist a topological sequence can be determined, but does not hinder the topological ordering
Code:
<span style= "FONT-SIZE:14PX;" >/* the adjacency Matrix O (n*n) node has been accessed, 0 means no, 1 indicates that it has been accessed, 1 means that it is being accessed, that is, in the recursive call frame */int g[n][n];int visit[n];int N; /*n indicates the number of nodes */bool dfs (int pos) {Visit[pos] =-1; for (int i=1; i <= n; i++) {if (G[pos][i]) {if (Visit[i] < 0) return false; else if (!visit[i] &&!dfs (i)) return false; }} Visit[pos] = 1; S.push (POS); return true;} /* The adjacency table O (n*e) node has been accessed, 0 means no, 1 indicates that it has been accessed, 1 means that it is being accessed, that is, in the recursive call frame */#include <bits/stdc++.h>using namespace Std;const int n=1e5+100;vector<int> g[n];int vis[n];bool dfs (int u) {vis[u]=-1; for (int i=0; i<g[u].size (); i++) {int v=g[u][i]; if (vis[v]<0) return false; else if (!vis[v]&&!dfs (v)) return false; } vis[u]=1; return true;} BOOL Topu (int n) {for (int i=1; i<=n; i++) if (!vis[i]) if (!dfs (i)) return false; return true;} int main () {int t; scanf ("%d", &t); while (t--){int n,m,v,u; memset (vis,0,sizeof (VIS)); scanf ("%d%d", &n,&m); for (int i=0; i<=n; i++) g[i].clear (); for (int i=0; i<m; i++) {scanf ("%d%d", &v,&u); G[v].push_back (U); } if (Topu (n)) puts ("YES"); Else puts ("NO"); } return 0;} </span>
Topological sequencing and continuation