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PT07Y - Is it a tree |
You are given an unweighted, undirected graph. Write a program to check if it's a tree topology.
Input
The first line of the input file contains two integers N and M --- number of nodes and number of edges in the graph (0 < N <= 10000, 0 <= M <= 20000). Next M lines contain M edges of that graph --- Each line contains a pair (u, v) means there is an edge between node u and node v (1 <= u, v <= N).
Output
Print YES if the given graph is a tree, otherwise print NO.
Example
Input: 3 2 1 2 2 3 Output: YES
| Added by: | Thanh-Vy Hua |
| Date: | 2007-03-28 |
| Time limit: | 0.5s |
| Source limit: | 50000B |
| Memory limit: | 1536MB |
| Cluster: | Cube (Intel G860) |
| Languages: | All except: ERL JS-RHINO |
| Resource: | Co-author Amber |
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2017-12-16 15:23:56
AC in one Go!!! |
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2017-12-12 15:58:49
AC in one GO!! |
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2017-11-14 08:15:54
My 50th!! AC in one Go!! Solved using Union-Find |
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2017-10-26 14:36:12
java people use input method other than scanner class it gives me one tle. just use Adjacency List and dfs. |
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2017-10-22 10:24:57
It can also also be solved as follow: Just check whether both edges you are taking as input is already present or not. |
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2017-10-19 13:02:25
Just Disjoint Set works quite well |
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2017-10-03 17:53:33
cool problem |
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2017-08-27 16:59:31
what if test cases contain self loops ?? even if we run a dfs and check if(m==n-1&&all nodes are visited) ..we print yes bt actually it should be no ri8 ?? can anyone clarify dis ?? |
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2017-08-14 23:38:02
Run dfs only once. Then check whether all the vertices are visited and no of edges equals no of vertices minus one. |
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2017-08-14 13:10:27
bfs or dfs or union find.But you have to consider the disconnected graphs also, which are not trees. |
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