A-Z Graph
You are given a directed graph consisting of $$$n$$$ vertices. Each directed edge (or arc) labeled with a single character. Initially, the graph is empty.You should process $$$m$$$ queries with it. Each query is one of three types:   "$$$+$$$ $$$u$$$ $$$v$$$ $$$c$$$" — add arc from $$$u$$$ to $$$v$$$ with label $$$c$$$. It's guaranteed that there is no arc $$$(u, v)$$$ in the graph at this moment;  "$$$-$$$ $$$u$$$ $$$v$$$" — erase arc from $$$u$$$ to $$$v$$$. It's guaranteed that the graph contains arc $$$(u, v)$$$ at this moment;  "$$$?$$$ $$$k$$$" — find the sequence of $$$k$$$ vertices $$$v_1, v_2, \dots, v_k$$$ such that there exist both routes $$$v_1 \to v_2 \to \dots \to v_k$$$ and $$$v_k \to v_{k - 1} \to \dots \to v_1$$$ and if you write down characters along both routes you'll get the same string. You can visit the same vertices any number of times. 
In the first query of the third type $$$k = 3$$$, we can, for example, choose a sequence $$$[1, 2, 3]$$$, since $$$1 \xrightarrow{\text{a}} 2 \xrightarrow{\text{b}} 3$$$ and $$$3 \xrightarrow{\text{a}} 2 \xrightarrow{\text{b}} 1$$$.In the second query of the third type $$$k = 2$$$, and we can't find sequence $$$p_1, p_2$$$ such that arcs $$$(p_1, p_2)$$$ and $$$(p_2, p_1)$$$ have the same characters.In the third query of the third type, we can, for example, choose a sequence $$$[1, 2, 3, 2, 1]$$$, where $$$1 \xrightarrow{\text{a}} 2 \xrightarrow{\text{b}} 3 \xrightarrow{\text{d}} 2 \xrightarrow{\text{c}} 1$$$.
The first line contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 2 \cdot 10^5$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of vertices in the graph and the number of queries.The next $$$m$$$ lines contain queries — one per line. Each query is one of three types:   "$$$+$$$ $$$u$$$ $$$v$$$ $$$c$$$" ($$$1 \le u, v \le n$$$; $$$u \neq v$$$; $$$c$$$ is a lowercase Latin letter);  "$$$-$$$ $$$u$$$ $$$v$$$" ($$$1 \le u, v \le n$$$; $$$u \neq v$$$);  "$$$?$$$ $$$k$$$" ($$$2 \le k \le 10^5$$$). It's guaranteed that you don't add multiple edges and erase only existing edges. Also, there is at least one query of the third type.
For each query of the third type, print YES if there exist the sequence $$$v_1, v_2, \dots, v_k$$$ described above, or NO otherwise.
import sys
input = sys.stdin.readline

def solve():
    (n, m) = map(int, input().split())
    e = dict()
    same = 0
    diff = 0
    res = []
    for i in range(m):
        a = input().split()
        c = a[0]
        if c == '+':
            u = int(a[1])
            v = int(a[2])
            c = a[3]
            z = e.get((v, u), None)
            if z == c:
                same += 1
            elif z is not None:
                diff += 1
            e[u, v] = c
        elif c == '-':
            u = int(a[1])
            v = int(a[2])
            c = e[u, v]
            del e[u, v]
            z = e.get((v, u), None)
            if z == c:
                same -= 1
            elif z is not None:
                diff -= 1
        else:
            c = int(a[1])
            if c % 2 == 1:
                if diff + same > 0:
                    res.append('YES')
                else:
                    res.append('NO')
            elif same > 0:
                res.append('YES')
            else:
                res.append('NO')
    print('\n'.join(res))
solve()
def solve():
    (n, m) = map(int, input().split())
    e = dict()
    same = 0
    diff = 0
    res = []
    for i in range(m):
        a = input().split()
        c = a[0]
        if c == '+':
            u = int(a[1])
            v = int(a[2])
            c = a[3]
            z = e.get((v, u), None)
            if z == c:
                same += 1
            elif z is not None:
                diff += 1
            e[u, v] = c
        elif c == '-':
            u = int(a[1])
            v = int(a[2])
            c = e[u, v]
            del e[u, v]
            z = e.get((v, u), None)
            if z == c:
                same -= 1
            elif z is not None:
                diff -= 1
        else:
            c = int(a[1])
            if c % 2 == 1:
                if diff + same > 0:
                    res.append('YES')
                else:
                    res.append('NO')
            elif same > 0:
                res.append('YES')
            else:
                res.append('NO')
    print('\n'.join(res))
solve()
def solve():
    (n, m) = map(int, input().split())
    e = dict()
    same = 0
    diff = 0
    res = []
    for i in range(m):
        a = input().split()
        c = a[0]
        if c == '+':
            u = int(a[1])
            v = int(a[2])
            c = a[3]
            z = e.get((v, u), None)
            if z == c:
                same += 1
            elif z is not None:
                diff += 1
            e[u, v] = c
        elif c == '-':
            u = int(a[1])
            v = int(a[2])
            c = e[u, v]
            del e[u, v]
            z = e.get((v, u), None)
            if z == c:
                same -= 1
            elif z is not None:
                diff -= 1
        else:
            c = int(a[1])
            if c % 2 == 1:
                res.append('YES' if diff + same > 0 else 'NO')
            else:
                res.append('YES' if same > 0 else 'NO')
    print('\n'.join(res))
solve()