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

hide comments
le_louch007: 2024-04-18 01:37:25

hint : a node cannot have 2 parents

mushfiq123: 2024-02-11 17:06:51

should have added a test case with disconnected components

k124k3n: 2023-08-19 06:23:21

bad problem. Maybe there's duplicate edge in problem like:
{1, 2}
{2, 1}

E==N-1 property maybe not happening but still a Tree

Last edit: 2023-08-19 06:24:06
zeyf: 2022-04-11 00:51:23

You can just use a UnionFind/DisjointSet for all other cases and whenever there is a failed union (returns false), this is the case where both nodes u and v are part of the same set already. This means it is not a tree in it's current form!

amansingh_20: 2021-11-29 09:25:50

AC in one go....
Just check cycle and connected components.

shofiqur_052: 2021-08-17 21:50:21

Number of connected component 1
And M < N ... Enough to get AC;

avx_5801: 2021-07-09 11:08:30

You have to also check self loop condition,then you get AC in one GO by Using simple dfs :)

mortal_beast: 2021-06-30 15:19:14

easy one

the_art_maniac: 2021-05-29 00:17:25

Some test cases that will help you:
9 8
7 8
2 3
8 3
6 3
5 3
7 9
9 1
9 4
YES
9 8
1 2
2 3
3 4
1 4
5 6
7 8
9 7
8 9
NO

yasser1110: 2021-02-19 14:54:05

+1 @jrseinc disjoint union set solution is the most straightforward one. Even though adjacency matrix and adjacency list solutions also work.

Last edit: 2021-02-19 14:54:25

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