The Majestic Brown Tree Snake (Codeforces 1381D)
Tags: Greedy, trees

There is an undirected tree of $$$n$$$ vertices, connected by $$$n-1$$$ bidirectional edges. There is also a snake stuck inside of this tree. Its head is at vertex $$$a$$$ and its tail is at vertex $$$b$$$. The snake's body occupies all vertices on the unique simple path between $$$a$$$ and $$$b$$$.The snake wants to know if it can reverse itself  — that is, to move its head to where its tail started, and its tail to where its head started. Unfortunately, the snake's movements are restricted to the tree's structure.In an operation, the snake can move its head to an adjacent vertex not currently occupied by the snake. When it does this, the tail moves one vertex closer to the head, so that the length of the snake remains unchanged. Similarly, the snake can also move its tail to an adjacent vertex not currently occupied by the snake. When it does this, the head moves one unit closer to the tail.  Let's denote a snake position by $$$(h,t)$$$, where $$$h$$$ is the index of the vertex with the snake's head, $$$t$$$ is the index of the vertex with the snake's tail. This snake can reverse itself with the movements $$$(4,7)\to (5,1)\to (4,2)\to (1, 3)\to (7,2)\to (8,1)\to (7,4)$$$. Determine if it is possible to reverse the snake with some sequence of operations.

Input: The first line contains a single integer $$$t$$$ ($$$1\le t\le 100$$$)  — the number of test cases. The next lines contain descriptions of test cases.The first line of each test case contains three integers $$$n,a,b$$$ ($$$2\le n\le 10^5,1\le a,b\le n,a\ne b$$$).Each of the next $$$n-1$$$ lines contains two integers $$$u_i,v_i$$$ ($$$1\le u_i,v_i\le n,u_i\ne v_i$$$), indicating an edge between vertices $$$u_i$$$ and $$$v_i$$$. It is guaranteed that the given edges form a tree.It is guaranteed that the sum of $$$n$$$ across all test cases does not exceed $$$10^5$$$.

Output: For each test case, output "YES" if it is possible for the snake to reverse itself, or "NO" otherwise.

Note: The first test case is pictured above.In the second test case, the tree is a path. We can show that the snake cannot reverse itself.In the third test case, we can show that the snake cannot reverse itself.In the fourth test case, an example solution is:$$$(15,12)\to (16,11)\to (15,13)\to (10,14)\to (8,13)\to (4,11)\to (1,10)$$$$$$\to (2,8)\to (3,4)\to (2,5)\to (1,6)\to (4,7)\to (8,6)\to (10,5)$$$$$$\to (11,4)\to (13,8)\to (14,10)\to (13,15)\to (11,16)\to (12,15)$$$.

Sample Inputs:
4
8 4 7
1 2
2 3
1 4
4 5
4 6
1 7
7 8
4 3 2
4 3
1 2
2 3
9 3 5
1 2
2 3
3 4
1 5
5 6
6 7
1 8
8 9
16 15 12
1 2
2 3
1 4
4 5
5 6
6 7
4 8
8 9
8 10
10 11
11 12
11 13
13 14
10 15
15 16

Sample Outputs:
YES
NO
NO
YES

Solution (Python):
import sys
from collections import deque
t = int(input())
for _ in range(t):
    (n, a, b) = [int(x) for x in input().split()]
    a -= 1
    b -= 1
    edges = set()
    adj = [[] for x in range(n)]
    for _ in range(n - 1):
        (u, v) = [int(x) for x in sys.stdin.readline().split()]
        u -= 1
        v -= 1
        edges.add((u, v))
        edges.add((v, u))
        adj[u].append(v)
        adj[v].append(u)
    to_a = [-1 for x in range(n)]
    to_a[a] = a
    stack = [a]
    while len(stack):
        cur = stack.pop()
        for nb in adj[cur]:
            if to_a[nb] == -1:
                to_a[nb] = cur
                stack.append(nb)
    snake = [b]
    while snake[-1] != a:
        snake.append(to_a[snake[-1]])
    snake = deque(snake)
    adj = [set(l) for l in adj]
    leaves = [x for x in range(n) if len(adj[x]) == 1]
    num_branch_points = sum([1 for l in adj if len(l) >= 3])
    new_leaves = []
    if len(snake) == 2:
        print('YES' if num_branch_points >= 1 else 'NO')
        continue
    while True:
        head = snake.pop()
        tail = snake.popleft()
        if len(adj[head]) == 1 and len(adj[tail]) == 1:
            print('NO')
            break
        if len(adj[head]) != 1:
            snake.append(head)
        else:
            snake.appendleft(tail)
        for leaf in leaves:
            if len(adj[leaf]) == 0:
                continue
            nb = adj[leaf].pop()
            adj[nb].remove(leaf)
            if len(adj[nb]) == 2:
                num_branch_points -= 1
            if len(adj[nb]) == 1:
                new_leaves.append(nb)
        (leaves, new_leaves) = (new_leaves, [])
        if num_branch_points == 0:
            print('NO')
            break
        if len(snake) == 2:
            print('YES')
            break

Solution (Python):
import sys
from collections import deque
t = int(input())
for _ in range(t):
    (n, a, b) = [int(x) for x in input().split()]
    a -= 1
    b -= 1
    edges = set()
    adj = [[] for x in range(n)]
    for _ in range(n - 1):
        (u, v) = [int(x) for x in sys.stdin.readline().split()]
        u -= 1
        v -= 1
        edges.add((u, v))
        edges.add((v, u))
        adj[u].append(v)
        adj[v].append(u)
    to_a = [-1 for x in range(n)]
    to_a[a] = a
    stack = [a]
    while len(stack):
        cur = stack.pop()
        for nb in adj[cur]:
            if to_a[nb] == -1:
                to_a[nb] = cur
                stack.append(nb)
    snake = [b]
    while snake[-1] != a:
        snake.append(to_a[snake[-1]])
    snake = deque(snake)
    adj = [set(l) for l in adj]
    leaves = [x for x in range(n) if len(adj[x]) == 1]
    num_branch_points = sum([1 for l in adj if len(l) >= 3])
    new_leaves = []
    if len(snake) == 2:
        print('YES' if num_branch_points >= 1 else 'NO')
        continue
    while True:
        head = snake.pop()
        tail = snake.popleft()
        if len(adj[head]) == 1 and len(adj[tail]) == 1:
            print('NO')
            break
        if len(adj[head]) != 1:
            snake.append(head)
        else:
            snake.appendleft(tail)
        for leaf in leaves:
            if len(adj[leaf]) == 0:
                continue
            nb = adj[leaf].pop()
            adj[nb].remove(leaf)
            if len(adj[nb]) == 2:
                num_branch_points -= 1
            if len(adj[nb]) == 1:
                new_leaves.append(nb)
        (leaves, new_leaves) = (new_leaves, [])
        if num_branch_points == 0:
            print('NO')
            break
        if len(snake) == 2:
            print('YES')
            break

Solution (Python):
class Solution:

    def __init__(self):
        self.his = set()
        self.dp = {}
        self.get_input()

    def get_input(self):
        (self.n, head, tail) = [int(tmp) for tmp in input().strip().split(' ')]
        head -= 1
        tail -= 1
        self.graph = [[] for _ in range(self.n)]
        for _ in range(self.n - 1):
            (x, y) = [int(tmp) - 1 for tmp in input().strip().split(' ')]
            self.graph[x].append(y)
            self.graph[y].append(x)
        pre = [-1] * self.n
        length = [-1] * self.n
        length[head] = 1
        lst_a = [head]
        lst_b = []
        while length[tail] < 0:
            for x in lst_a:
                for y in self.graph[x]:
                    if length[y] >= 0:
                        continue
                    pre[y] = x
                    length[y] = length[x] + 1
                    lst_b.append(y)
            lst_a = lst_b
            lst_b = []
        self.snake = [tail]
        while self.snake[-1] != head:
            self.snake.append(pre[self.snake[-1]])

    def get_path(self):
        (self.path, self.sec) = self.get_path_rec(self.snake[1], self.snake[0])

    def get_path_rec(self, a, b):
        oa = a
        ob = b
        if (a, b) in self.dp:
            return self.dp[a, b]
        pre_path = [b]
        while len(self.graph[b]) == 2:
            tmp = b
            b = self.graph[b][0] if self.graph[b][0] != a else self.graph[b][1]
            a = tmp
            pre_path = [b] + pre_path
        if len(self.graph[b]) == 1:
            self.dp[oa, ob] = (pre_path, 0)
            return (pre_path, 0)
        path_list = []
        sec_list = []
        for x in self.graph[b]:
            if x == a:
                continue
            (path, sec) = self.get_path_rec(b, x)
            path_list.append(path)
            sec_list.append(sec)
        path_list.sort(key=lambda x: -len(x))
        self.dp[oa, ob] = (path_list[0] + pre_path, max(len(path_list[1]) + 1, max(sec_list)))
        return (path_list[0] + pre_path, max(len(path_list[1]) + 1, max(sec_list)))

    def move(self):
        length = len(self.snake)
        self.snake = self.path[:-1] + self.snake
        self.snake = self.snake[:length]
        for i in range(int(len(self.snake) / 2)):
            (self.snake[i], self.snake[-i - 1]) = (self.snake[-i - 1], self.snake[i])

    def get_result(self):
        if len(self.snake) == 1:
            return 'YES'
        while (self.snake[1], self.snake[0]) not in self.his:
            self.his.add((self.snake[1], self.snake[0]))
            self.get_path()
            if self.sec >= len(self.snake):
                return 'YES'
            self.move()
        return 'NO'
if __name__ == '__main__':
    t = int(input().strip())
    for _ in range(t):
        s = Solution()
        print(s.get_result())