Problem:


Score : 500 points

Problem Statement

Takahashi is not good at problems about trees in programming contests, and Aoki is helping him practice.

First, Takahashi created a tree with N vertices numbered 1 through N, and wrote 0 at each edge.

Then, Aoki gave him M queries. The i-th of them is as follows:

Increment the number written at each edge along the path connecting vertices a_i and b_i, by one.

After Takahashi executed all of the queries, he told Aoki that, for every edge, the written number became an even number.
However, Aoki forgot to confirm that the graph Takahashi created was actually a tree, and it is possible that Takahashi made a mistake in creating a tree or executing queries.

Determine whether there exists a tree that has the property mentioned by Takahashi.

Constraints

2 ≤ N ≤ 10^5

1 ≤ M ≤ 10^5

1 ≤ a_i,b_i ≤ N

a_i ≠ b_i

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
:
a_M b_M

Output

Print YES if there exists a tree that has the property mentioned by Takahashi; print NO otherwise.

Sample Input 1

4 4
1 2
2 4
1 3
3 4

Sample Output 1

YES

For example, Takahashi's graph has the property mentioned by him if it has the following edges: 1-2, 1-3 and 1-4.
In this case, the number written at every edge will become 2.

Sample Input 2

5 5
1 2
3 5
5 1
3 4
2 3

Sample Output 2

NO



Sample input:
4 4
1 2
2 4
1 3
3 4


Sample output:
YES

Submission status: Accepted

Solution code:
def main(sample_file = ''):

    """ convenient functions
    # for i, a in enumerate(iterable)
    # q, mod = divmod(a, b)
    # divmod(x, y) returns the tuple (x//y, x%y)
    # Higher-order function: reduce(operator.mul, xyz_count, 1)
    # manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
    """

    """convenient decorator
    # @functools.lru_cache():
    # to facilitate use of recursive function
        # ex:
        # from functools import lru_cache
        # import sys
        # sys.setrecursionlimit(10**9)
        # @lru_cache(maxsize=None)
        # def fib(n):
        #     if n < 2:
        #         return n
        #     return fib(n-1) + fib(n-2)
        # print(fib(1000))
    """
    
    # import numpy as np
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product # https://docs.python.org/ja/3/library/itertools.html
    # accumulate() returns iterator! to get list: list(accumulate())
    from math import factorial, ceil, floor, sqrt
    def factorize(n):
        """return the factors of the Arg and count of each factor
        
        Args:
            n (long): number to be resolved into factors
        
        Returns:
            list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
        """
        fct = []  # prime factor
        b, e = 2, 0  # base, exponent
        while b * b <= n:
            while n % b == 0:
                n = n // b
                e = e + 1
            if e > 0:
                fct.append((b, e))
            b, e = b + 1, 0
        if n > 1:
            fct.append((n, 1))
        return fct
    def combinations_count(n, r):   
        """Return the number of selecting r pieces of items from n kinds of items.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        # TODO: How should I do when n - r is negative?
        if n < 0 or r < 0:
            raise Exception('combinations_count(n, r) not defined when n or r is negative')
        if n - r < r: r = n - r
        if r < 0: return 0
        if r == 0: return 1
        if r == 1: return n
        numerator = [n - r + k + 1 for k in range(r)]
        denominator = [k + 1 for k in range(r)]
        for p in range(2,r+1):
            pivot = denominator[p - 1]
            if pivot > 1:
                offset = (n - r) % p
                for k in range(p-1,r,p):
                    numerator[k - offset] /= pivot
                    denominator[k] /= pivot
        result = 1
        for k in range(r):
            if numerator[k] > 1:
                result *= int(numerator[k])
        return result
    def combinations_with_replacement_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        if n < 0 or r < 0:
            raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
        elif n == 0:
            return 1
        else:
            return combinations_count(n + r - 1, r)
    from bisect import bisect_left, bisect_right
    from collections import deque, Counter, defaultdict # https://docs.python.org/ja/3/library/collections.html#collections.deque
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    from operator import itemgetter #sort 
    # ex1: List.sort(key=itemgetter(1))
    # ex2: sorted(tuples, key=itemgetter(1,2))
    from functools import reduce, lru_cache
    def chmin(x, y):
        """change minimum
        if x > y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current minimum value
            y (long): potential minimum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x > y:
            x = y
            return (x, True)
        else:
            return (x, False)
    def chmax(x, y):
        """change maximum
        if x < y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current maximum value
            y (long): potential maximum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x < y:
            x = y
            return (x, True)
        else:
            return (x, False)

    from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
    def gcds(numbers):
        return reduce(gcd, numbers)
    def lcm(x, y):
        return (x * y) // gcd(x, y)
    def lcms(numbers):
        return reduce(lcm, numbers, 1)
    def make_divisors(n, reversed=False):
        """create list of divisors
        
        Args:
            number (int): number from which list of divisors is created
            reversed (bool, optional): ascending order if False. Defaults to False.
        
        Returns:
            list: list of divisors
        """
        divisors = set()
        for i in range(1, int(n**0.5)+1):
            if n % i == 0:
                divisors.add(i)
                divisors.add(n//i)
        return sorted(list(divisors),reverse=reversed)
    
    # first create factorial_list
    # fac_list = mod_factorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    AtoZ = list('abcdefghijklmnopqrsruvwxyz')
    modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
    def modinv(a, p = MOD):
        # evaluate reciprocal using Fermat's little theorem:
        # a**(p-1) is identical to 1 (mod p) when a and p is coprime
        return modpow(a, p-2, p)
    def modinv_list(n, p = MOD):
        if n <= 1:
            return [0,1][:n+1]
        else:
            inv_t = [0,1]
            for i in range(2, n+1):
                inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
            return inv_t
    def modfactorial_list(n, p = MOD):
        if n == 0:
            return [1]
        else:
            l = [0] * (n+1)
            tmp = 1
            for i in range(1, n+1):
                tmp = tmp * i % p
                l[i] = tmp
            return l
    def modcomb(n, k, fac_list = [], p = MOD):
        # fac_list = modfactorial_list(100)
        # print(modcomb(100, 5, modfactorial_list(100)))
        from math import factorial
        if n < 0 or k < 0 or n < k: return 0
        if n == 0 or k == 0: return 1
        if len(fac_list) <= n:
            a = factorial(n) % p
            b = factorial(k) % p
            c = factorial(n-k) % p
        else:
            a = fac_list[n]
            b = fac_list[k]
            c = fac_list[n-k]
        return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
    def modadd(a, b, p = MOD):
        return (a + b) % MOD
    def modsub(a, b, p = MOD):
        return (a - b) % p
    def modmul(a, b, p = MOD):
        return ((a % p) * (b % p)) % p
    def moddiv(a, b, p = MOD):
        return modmul(a, modpow(b, p-2, p))

    class UnionFindTree:
        """union find tree class
        TODO: fix this description...
        how to use (example):
        >>  uf = UnionFindTree(N) 
        >>  if uf.find_root(a) == uf.find_root(b):
        >>      do something
        >>  else:
        >>      do something
        >>      uf.unite(a, b)
        """
        def __init__(self, N):
            self.root = [-1] * (N+1)
            self.rank   = [0]  * (N+1)
            self.connected_num = [1] * (N+1)
    
        def find_root(self,x):
            root = self.root
            while root[x] != -1:
                x = root[x]
            return x
    
        def unite(self,x,y):
            root = self.root
            rank = self.rank
            connected_num = self.connected_num
            find_root = self.find_root
    
            rx = find_root(x)
            ry = find_root(y)
            if rx != ry:
                if rank[rx] < rank[ry]:
                    root[rx] = ry
                    rx,ry = ry,rx
                else:
                    if rank[rx] == rank[ry]:
                        rank[rx] += 1
                    root[ry] = rx       
                connected_num[rx] += connected_num[ry]

    # ************ Graph ************
    # Graph: https://en.wikipedia.org/wiki/Directed_graph
    # 最短経路問題: https://ja.wikipedia.org/wiki/%E6%9C%80%E7%9F%AD%E7%B5%8C%E8%B7%AF%E5%95%8F%E9%A1%8C
    
    # Bellman-Ford: O(|V||E|). Use this if there exists an edge with negative length in the graph
    # After N steps, the shortest path has converded if there doesn't exist an cycle of edges with negative
    # Watch out: d[N] == d[2*N] doesn't necessarily mean the graph doesn't have negative cycle
    # ref: https://www.youtube.com/watch?v=1Z6ofKN03_Y

    def BellmanFord(N, M, ABC, vertex_start, vertex_end, value_if_inf = -1, find_shortest = False):
        """to calculate furthest or shortest length between vertex_start and vertex_end using BellmanFord algorithm
        
        Args:
            N (int): number of vertices
            M (int): number of edges
            ABC (list): [(ai, bi, ci) for _ in range(N)] where i-th edge is directed from vertex ai to vertex bi and the length is ci 
            vertex_start (int): start vertex. usually use 0.
            vertex_end (int): end vertex. usually use N-1.
            value_if_inf (int or string as you like, optional): value you want when the furthest (or shortest) distance is infinite (or -infinite). Defaults to -1.
            find_shortest (bool, optional): choose False to find furthest path. Defaults to False.
        
        Returns:
            int or string: normally int (but can be str if you set value_if_inf to str)
            
        Example:
                N, M, P = R()
                ABC = [R() for _ in range(M)]
                ABC = [(a-1, b-1, c-P) for a, b, c in ABC]
                print(BellmanFord(N, M, ABC, 0, N-1, value_if_inf = 'inf'))

        """

        def make_reachable_list(N, M, ABC, vertex_start, vertex_end):
            reachable_to_direct = defaultdict(list)
            reachable_from_direct = defaultdict(list)
            reachable_from_start = [False] * N
            reachable_to_end = [False] * N
            reachable_from_start[vertex_start] = True
            reachable_to_end[vertex_end] = True
            reachable_from_both_sides = [False] * N
            dfs_from_start = []
            dfs_to_end = []
            for a, b, c in ABC:
                reachable_to_direct[a].append(b)
                reachable_from_direct[b].append(a)
                if a == vertex_start:
                    dfs_from_start.append(b)
                    reachable_from_start[b] = True
                if b == vertex_end:
                    dfs_to_end.append(a)
                    reachable_to_end[a] = True
            while dfs_from_start:
                v = dfs_from_start.pop()
                for i in reachable_to_direct[v]:
                    if not reachable_from_start[i]:
                        reachable_from_start[i] = True
                        dfs_from_start.append(i)
            while dfs_to_end:
                v = dfs_to_end.pop()
                for i in reachable_from_direct[v]:
                    if not reachable_to_end[i]:
                        reachable_to_end[i] = True
                        dfs_to_end.append(i)
            for i in range(N):
                reachable_from_both_sides[i] = reachable_from_start[i] and reachable_to_end[i]
            return reachable_from_both_sides
        
        reachable_from_both_sides = make_reachable_list(N, M, ABC, vertex_start, vertex_end)
        
        if find_shortest:
            dist = [INF for i in range(N)]
        else:
            dist = [-INF for i in range(N)]
        
        dist[vertex_start] = 0
        for i in range(N):
            updated = False
            for a, b, c in ABC:
                if not reachable_from_both_sides[a]:
                    continue
                elif find_shortest:
                    update_condition = dist[a] + c < dist[b]
                else:
                    update_condition = dist[a] + c > dist[b]
                if dist[a] != INF and update_condition:
                    dist[b] = dist[a] + c
                    updated = True
                    if i == N-1:
                        return value_if_inf
            if not updated:
                break
        return dist[vertex_end]

    # Warshall Floyd O(V**3)
    def warshall_floyd(number_vertex, XYD, shift_one_for_vertex=False):
        # ref: https://juppy.hatenablog.com/entry/2018/11/01/%E8%9F%BB%E6%9C%AC_python_%E5%85%A8%E7%82%B9%E5%AF%BE%E6%9C%80%E7%9F%AD%E7%B5%8C%E8%B7%AF%E6%B3%95%EF%BC%88%E3%83%AF%E3%83%BC%E3%82%B7%E3%83%A3%E3%83%AB%E3%83%95%E3%83%AD%E3%82%A4%E3%83%89%E6%B3%95
        # number_vertex: len(vertex)
        # XYD: list of distance from x to y
        # shift_one_for_vertex: XYDの値が1~Nの時(0~N-1でない時)に調整するフラグ。
        # d[i][j]: iからjへの最短距離
        d = [[float("inf")]*number_vertex for i in range(number_vertex)]
        for x, y, dist in XYD:
            if shift_one_for_vertex:
                x = x - 1
                y = y - 1
            d[x][y] = dist
            d[y][x] = dist
        for i in range(number_vertex):
            d[i][i] = 0 # distance of same vertex is 0
        for k in range(number_vertex):
            for i in range(number_vertex):
                for j in range(number_vertex):
                    d[i][j] = min(d[i][j],d[i][k] + d[k][j])
        return d


    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array:
            # ex) [[0] * N for _ in range(N)]
            # ex2) dp = [[0] * (N+1) for _ in range(W*2)]
    # set inputs
        # put inputs between specific values (almost as quickly)
        # ex) S = [-INF] + [int(r()) for _ in range(A)] + [INF]
        # open(0).read() is sometimes useful:
        # ex) n, m, *x = map(int, open(0).read().split())
        #     min(x[::2]) - max(x[1::2])
        # ex2) *x, = map(int, open(0).read().split())
        #     don't forget to add comma after *x if only one variable is used
    # preprocessing
        # transpose = [x for x in zip(*data)]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
        # flat = [flatten for inner in data for flatten in inner]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [1, 2, 3, 4, 5, 6, 7, 8, 9]
    # calculate and output
        # output pattern
        # ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
    """

    # functions to read input
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    r_float = lambda: float(r())
    R = lambda: list(map(int, r().split()))
    R_str = lambda: r().split()
    R_map = lambda: map(int, r().split())
    R_float = lambda: list(map(float, r().split()))
    R_tuple = lambda: tuple(map(int, r().split()))

    """ how to treat input
    # single int: int(r())
    # single string: r()
    # single float: float(r())
    # line int: R()
    # line string: r().split()
    # line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
    # lines int: [R() for _ in range(n)]
    """

    # for test
    if sample_file:
        sys.stdin = open(sample_file)
        
    # ----------------------------------
    # main
    #%%
    N, M = R()
    AB = [R() for i in range(M)]
    l = [0] * N
    for a, b in AB:
        l[a-1] += 1
        l[b-1] += 1
    flag = True
    for ll in l:
        if ll%2 != 0:
            flag = False
    if flag:
        print('YES')
    else:
        print('NO')

    #%%
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # three methods for loop: keys(), values(), items()
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    # to make binary digits text
    >>> a = 5
    >>> bin_str_a = format(a, '#06b')
    >>> print(bin_str_a)
    0b0101 # first 2 strings (='0b') indicates it is binary
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
n, m = map(int, input().split())

tree = [0 for _ in range(n+1)]
for _i in range(m):
    a, b = map(int, input().split())
    tree[a] += 1
    tree[b] += 1
if all(i%2==0 for i in tree):
    print("YES")
else:
    print("NO")

Submission status: Accepted

Solution code:
#AGC014-B Unplanned Queries
"""
今回は各辺の偶奇のみを考えるので、
根付き木を考えた時に、頂点a,bから親側に全部流した時に、
LCAより先は勝手に偶数になるので無視できる。
とした場合に、根をrootとして、各クエリに対して
a-root,b-rootへのパスに1を足すことと考えると、
結局各辺の偶奇はaとbの出現回数の偶奇によると考える事ができる。
よって、クエリに与えられる辺の回数をdictで保持し、
最終的に一つでも奇数回出現する頂点があるならばFalse
そうでないならTrueとなる

"""
import sys
from collections import defaultdict
readline = sys.stdin.buffer.readline
def even(n): return 1 if n%2==0 else 0
n,m = map(int,readline().split())
dic1 = defaultdict(int)
for i in range(m):
    a,b = map(int,readline().split())
    dic1[a] += 1
    dic1[b] += 1

for i in dic1.values():
    if not even(i):
        print("NO")
        exit()
print("YES")

Submission status: Accepted

Solution code:
#AGC014-B Unplanned Queries
import sys
from collections import defaultdict
readline = sys.stdin.buffer.readline
def even(n): return 1 if n%2==0 else 0
n,m = map(int,readline().split())
dic1 = defaultdict(int)
for i in range(m):
    a,b = map(int,readline().split())
    dic1[a] += 1
    dic1[b] += 1

for i in dic1.values():
    if not even(i):
        print("NO")
        exit()
print("YES")

Submission status: Accepted

Solution code:
N,M=map(int,input().split())

count=[0]*N
for _ in range(M):
    a,b=map(int,input().split())
    count[a-1]+=1
    count[b-1]+=1

for c in count:
    if c%2==1:
        print("NO")
        exit()

print("YES")

Submission status: Accepted

Solution code:
n,m = map(int,input().split())
ab = [[int(i) for i in input().split()] for j in range(m)]

im_b = [0]*(n)

for i in range(m):
    im_b[ab[i][0]-1] += 1
    im_b[ab[i][1]-1] -= 1

if im_b[0] % 2 == 1:
    print('NO')
    exit()

#im_a = [0]*(n)
#im_a[0] = im_b[0]

for i in range(n-1):
    #print(i)
    if im_b[i] % 2 == 1:
        print('NO')
        exit()

print('YES')

Submission status: Accepted

Solution code:
c = [0]*10**5
for i in open(0).read().split()[2:]: c[int(i)-1] ^= 1
print("NO" if any(c) else "YES")

Submission status: Accepted

Solution code:
n,m = map(int, input().split())
ab=[list(map(int,input().split())) for i in range(m)]
cnt=[0]*n

for tmp in ab:
    a,b=tmp
    cnt[a-1]+=1
    cnt[b-1]+=1

for c in cnt:
    if c%2!=0:
        print("NO")
        exit()
print("YES")

Submission status: Accepted

Solution code:
n,m = map(int,input().split())
from collections import Counter

s = []
for _ in range(m):
    a,b = map(int,input().split())
    s.append(a)
    s.append(b)
    
c = Counter(s)
d = list(c.values())

ans = "YES"
for i in d:
    if i % 2 == 1:
        ans = "NO"
        break
        
print(ans)

Submission status: Accepted

Solution code:
import sys
import math  # noqa
import bisect  # noqa
import queue  # noqa


def input():
    return sys.stdin.readline().rstrip()


def main():
    N, M = map(int, input().split())
    cnt = [0 for _ in range(N)]
    for _ in range(M):
        a, b = map(int, input().split())
        a -= 1
        b -= 1
        cnt[a] += 1
        cnt[b] += 1

    for c in cnt:
        if c % 2 != 0:
            return print("NO")

    print("YES")


if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
N,M=map(int, input().split())

D={} 
for i in range(N):
  D[i+1]=0
for i in range(M):
  a,b=map(int, input().split())
  D[a]+=1
  D[b]+=1

odds=0
evens=0
for v in D.values():
    if v%2==0:
        evens+=1
    else:
        odds+=1
if odds>0:
    print("NO")
else:
    print("YES")

Submission status: Accepted

Solution code:
N, M = map(int, input().split())
C = [0] * N
for i in range(M):
    ai, bi = map(int, input().split())
    C[ai - 1] += 1
    C[bi - 1] += 1

print('YES' if len(list(filter(lambda i: i % 2 == 1, C))) == 0 else 'NO')

Submission status: Accepted

Solution code:
c=[0]*9**6
for x in open(0).read().split()[2:]:c[int(x)]^=1
print("YNEOS"[any(c)::2])

Submission status: Accepted

Solution code:
n,m,*p=map(int,open(0).read().split());c=[0]*n
for x in p:c[x-1]+=1
print("YNEOS"[any(x%2for x in c)::2])

Submission status: Accepted

Solution code:
from collections import Counter

N, M = map(int, input().split())
counter = Counter()
for i in range(M):
    a, b = map(int, input().split())
    counter[a] += 1
    counter[b] += 1

if all(x % 2 == 0 for x in counter.values()):
    print("YES")
else:
    print("NO")

Submission status: Accepted

Solution code:
N,M=map(int,input().split())
G=[]
x=dict()
for i in range(M):
    a,b=map(int,input().split())
    if a in x:
        x[a]+=1
    else:
        x[a]=1
    if b in x:
        x[b]+=1
    else:
        x[b]=1

flag=True
for idx in x:
    if x[idx]%2==0:
        pass
    else:
        flag=False

if flag:
    print("YES")
else:
    print("NO")

Submission status: Accepted

Solution code:
import sys,collections as cl,bisect as bs
sys.setrecursionlimit(100000)
input = sys.stdin.readline
mod = 10**9+7
Max = sys.maxsize
def l(): #intのlist
    return list(map(int,input().split()))
def m(): #複数文字
    return map(int,input().split())
def onem(): #Nとかの取得
    return int(input())
def s(x): #圧縮
    a = []
    if len(x) == 0:
        return []
    aa = x[0]
    su = 1
    for i in range(len(x)-1):
        if aa != x[i+1]:
            a.append([aa,su])
            aa = x[i+1]
            su = 1
        else:
            su += 1
    a.append([aa,su])
    return a
def jo(x): #listをスペースごとに分ける
    return " ".join(map(str,x))
def max2(x): #他のときもどうように作成可能
    return max(map(max,x))
def In(x,a): #aがリスト(sorted)
    k = bs.bisect_left(a,x)
    if k != len(a) and a[k] ==  x:
        return True
    else:
        return False

def pow_k(x, n):
    ans = 1
    while n:
        if n % 2:
            ans *= x
        x *= x
        n >>= 1
    return ans

"""
def nibu(x,n,r):
    ll = 0
    rr = r
    while True:
        mid = (ll+rr)//2

    if rr == mid:
        return ll
    if (ここに評価入れる):
        rr = mid
    else:
        ll = mid+1
"""

n,k = m()

po = [0 for i in range(n)]

for i in range(k):
    a,b = m()
    po[a-1] += 1
    po[b-1] += 1
if all(map(lambda x:x%2 == 0,po)):
    print("YES")
else:
    print("NO")

Submission status: Accepted

Solution code:
N, M = map(int, input().split())
ab = []
for _ in range(M):
    ab.append(tuple(map(int, input().split())))

bucket = [0] * N
for el in ab:
    a, b = el
    bucket[a - 1] += 1
    bucket[b - 1] += 1
if all([x % 2 == 0 for x in bucket]):
    print('YES')
else:
    print('NO')

Submission status: Accepted

Solution code:
N, M = map(int, input().split())
ab = []
for _ in range(M):
    ab.append(tuple(map(int, input().split())))

edges = [0] * N
for el in ab:
    a = min(el)
    b = max(el)
    edges[a - 1] += 1
    edges[b - 1] -= 1
# print(edges)
for i in range(1, N):
    edges[i] = edges[i] + edges[i - 1]

# print(edges)
if all([x % 2 == 0 for x in edges]):
    print('YES')
else:
    print('NO')

Submission status: Accepted

Solution code:
n, m = map(int, input().split())
count = [0 for i in range(n+1)]
for i in range(m):
    a,b = map(int, input().split())
    count[a] += 1
    count[b] += 1
for i in range(n+1):
    if count[i] % 2 != 0:
        print("NO")
        exit()
print("YES")

Submission status: Accepted

Solution code:
from collections import defaultdict as d
N,M=map(int,input().split())
E=d(int)
for i in range(M):
  a,b=map(int,input().split())
  E[a]^=1
  E[b]^=1
print(("YES","NO")[max(E.values())])

Submission status: Accepted

Solution code:
# すべての頂点が入力に現れる回数が偶数であればOK 奇数回の頂点が一つでもあればNO
from collections import defaultdict as d

N,M = map(int,input().split())
E = d(int)

for i in range(M):
  a,b = map(int,input().split())
  E[a] ^= 1
  E[b] ^= 1

print(("YES","NO")[sorted(E.values())[-1]])

Submission status: Accepted

Solution code:
import sys
N, M = [int(i) for i in sys.stdin.readline().split()]
v = [0] * N
for _ in range(M):
    a, b = map(int, sys.stdin.readline().split())
    v[a - 1] ^= 1
    v[b - 1] ^= 1
print("NO" if 1 in v else "YES")

Submission status: Accepted

Solution code:
n,m=map(int,input().split())
a=[0]*(n+1)
for i in range(m):
  x,y=map(int,input().split())
  a[x]+=1
  a[y]+=1
ans='YES'
for i in range(1,n+1):
  if a[i]%2==1:
    ans='NO'
    break
print(ans)

Submission status: Accepted

Solution code:
# https://atcoder.jp/contests/agc014/tasks/agc014_b
# 条件を満たすものが存在するかの判別問題

# クエリをグラフにして考えてみる
# a-bを結ぶすべての経路は木上のa-b間の経路を必ず通る
# つまりa-b間の経路の数が偶数である必要がある。すべてのノードに対してエッジが偶数個つながっているかを確認すればよい

import sys
sys.setrecursionlimit(1 << 25)
read = sys.stdin.readline
ra = range
enu = enumerate


def read_ints():
    return list(map(int, read().split()))


N, M = read_ints()

n_edges = [0] * N
for _ in ra(M):
    a, b = read_ints()
    n_edges[a - 1] += 1
    n_edges[b - 1] += 1

for n in n_edges:
    if n & 1:
        print('NO')
        exit()
print('YES')

Submission status: Accepted

Solution code:
N, M = map(int, input().split())
cnt = [0] * (N+1)
for _ in range(M):
    ta,tb = map(int, input().split())
    cnt[ta] += 1
    cnt[tb] += 1
cond = 0
for i in range(1,N+1):
    if cnt[i] % 2 == 1:
        cond = 1
        break

if cond:
    print('NO')
else:
    print('YES')

Submission status: Accepted

Solution code:
#!/usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
from bisect import bisect_left, bisect_right
import sys, random, itertools, math
sys.setrecursionlimit(10**5)
input = sys.stdin.readline
sqrt = math.sqrt
def LI(): return list(map(int, input().split()))
def LF(): return list(map(float, input().split()))
def LI_(): return list(map(lambda x: int(x)-1, input().split()))
def II(): return int(input())
def IF(): return float(input())
def S(): return input().rstrip()
def LS(): return S().split()
def IR(n): return [II() for _ in range(n)]
def LIR(n): return [LI() for _ in range(n)]
def FR(n): return [IF() for _ in range(n)]
def LFR(n): return [LI() for _ in range(n)]
def LIR_(n): return [LI_() for _ in range(n)]
def SR(n): return [S() for _ in range(n)]
def LSR(n): return [LS() for _ in range(n)]
mod = 1000000007
inf = 1e10

#solve
def solve():
    n, m = LI()
    l = [0] * n
    for i in range(m):
        a, b = LI_()
        l[a] ^= 1
        l[b] ^= 1
    print("NO" if sum(l) else "YES")
    return


#main
if __name__ == '__main__':
    solve()

Submission status: Accepted

Solution code:
N,M = list(map(int, input().split()))
n = [0]*(N+1)
for _ in range(M):
    a,b = list(map(int, input().split()))
    n[a] += 1
    n[b] += 1


flag = True
for j in range(len(n)):
    if n[j]%2 != 0:
        flag = False

if flag:
    print('YES')
else:
    print('NO')

Submission status: Accepted

Solution code:
import sys
input = sys.stdin.readline
from collections import *

N, M = map(int, input().split())
dic = defaultdict(int)
for i in range(M):
    a, b = map(int, input().split())
    dic[a] += 1
    dic[b] += 1
if any([x&1 for x in dic.values()]):
    print('NO')
else:
    print('YES')

Submission status: Accepted

Solution code:
import sys, re
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians
from itertools import accumulate, permutations, combinations, product
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits
from bisect import bisect, bisect_left
from fractions import gcd
from heapq import heappush, heappop
from functools import reduce
def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
def ZIP(n): return zip(*(MAP() for _ in range(n)))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
mod = 10 ** 9 + 7

N, M = MAP()

a, b = ZIP(M)

a_cnt, b_cnt = Counter(a), Counter(b)

for i in range(1, N+1):
	if (a_cnt[i]+b_cnt[i])%2 != 0:
		print("NO")
		break
else:
	print("YES")

Submission status: Accepted

Solution code:
import sys, re
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians
from itertools import accumulate, permutations, combinations, product
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits
from bisect import bisect, bisect_left
from fractions import gcd
from heapq import heappush, heappop
from functools import reduce
def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
def ZIP(n): return zip(*(MAP() for _ in range(n)))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
mod = 10 ** 9 + 7

N, M = MAP()

a, b = ZIP(M)

a_cnt, b_cnt = Counter(a), Counter(b)

for i in range(N):
	if (a_cnt[i]+b_cnt[i])%2 != 0:
		print("NO")
		break
else:
	print("YES")

Submission status: Accepted

Solution code:
n,m=map(int,input().split())
AB=[list(map(int,input().split())) for i in range(m)]

Tree=[[] for _ in range(n+1)]
for a,b in AB:
    Tree[a].append(b)
    Tree[b].append(a)

for i in range(1,n+1):
    if len(Tree[i])%2==0:continue
    else:
        print("NO")
        exit()
print("YES")