Submit | All submissions | Best solutions | Back to list |
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 |
hide comments
|
||||||||||||||
2018-07-28 22:34:31
simple dfs. increment a counter on dfs function. if counter > node*node then there is a cycle. |
||||||||||||||
2018-06-24 23:04:46
Simple DFS!! |
||||||||||||||
2018-06-21 23:03:31
got two WA due..to yes instead of YES....finally AC |
||||||||||||||
2018-06-21 10:52:34
DSU all the way! |
||||||||||||||
2018-06-18 09:33:28
did bfs but used only simple arrays, comments helped a lot. Last edit: 2018-06-18 09:34:35 |
||||||||||||||
2018-06-11 22:16:04 Daniel Bezerra Galvão Mitre
WA some times for testcase 0 0 |
||||||||||||||
2018-06-09 20:03:04
accepted with tarjan's algorithm |
||||||||||||||
2018-06-08 10:00:24
No need DFS or DSU just a visited array works.... |
||||||||||||||
2018-06-05 13:32:30
Use #DSU AC in one go!!!! |
||||||||||||||
2018-04-06 17:31:22
DSU makes this super easy . #beware ( tot edges = tot nodes -1 , in a valid tree ) |