A14070. Christmas Game
编程题
普及/提高-
知识点
题目描述
Alice and Bob are going to celebrate Christmas by playing a game with a tree of presents. The tree has $n$ nodes (numbered $1$ to $n$ , with some node $r$ as its root). There are $a_i$ presents are hanging from the $i$ -th node.
Before beginning the game, a special integer $k$ is chosen. The game proceeds as follows:
- Alice begins the game, with moves alternating each turn;
- in any move, the current player may choose some node (for example, $i$ ) which has depth at least $k$ . Then, the player picks some positive number of presents hanging from that node, let's call it $m$ $(1 \le m \le a_i)$ ;
- the player then places these $m$ presents on the $k$ -th ancestor (let's call it $j$ ) of the $i$ -th node (the $k$ -th ancestor of vertex $i$ is a vertex $j$ such that $i$ is a descendant of $j$ , and the difference between the depth of $j$ and the depth of $i$ is exactly $k$ ). Now, the number of presents of the $i$ -th node $(a_i)$ is decreased by $m$ , and, correspondingly, $a_j$ is increased by $m$ ;
- Alice and Bob both play optimally. The player unable to make a move loses the game.
For each possible root of the tree, find who among Alice or Bob wins the game.
Note: The depth of a node $i$ in a tree with root $r$ is defined as the number of edges on the simple path from node $r$ to node $i$ . The depth of root $r$ itself is zero.
Before beginning the game, a special integer $k$ is chosen. The game proceeds as follows:
- Alice begins the game, with moves alternating each turn;
- in any move, the current player may choose some node (for example, $i$ ) which has depth at least $k$ . Then, the player picks some positive number of presents hanging from that node, let's call it $m$ $(1 \le m \le a_i)$ ;
- the player then places these $m$ presents on the $k$ -th ancestor (let's call it $j$ ) of the $i$ -th node (the $k$ -th ancestor of vertex $i$ is a vertex $j$ such that $i$ is a descendant of $j$ , and the difference between the depth of $j$ and the depth of $i$ is exactly $k$ ). Now, the number of presents of the $i$ -th node $(a_i)$ is decreased by $m$ , and, correspondingly, $a_j$ is increased by $m$ ;
- Alice and Bob both play optimally. The player unable to make a move loses the game.
For each possible root of the tree, find who among Alice or Bob wins the game.
Note: The depth of a node $i$ in a tree with root $r$ is defined as the number of edges on the simple path from node $r$ to node $i$ . The depth of root $r$ itself is zero.
输入格式
The first line contains two space-separated integers $n$ and $k$ $(3 \le n \le 10^5, 1 \le k \le 20)$ .
The next $n-1$ lines each contain two integers $x$ and $y$ $(1 \le x, y \le n, x \neq y)$ , denoting an undirected edge between the two nodes $x$ and $y$ . These edges form a tree of $n$ nodes.
The next line contains $n$ space-separated integers denoting the array $a$ $(0 \le a_i \le 10^9)$ .
The next $n-1$ lines each contain two integers $x$ and $y$ $(1 \le x, y \le n, x \neq y)$ , denoting an undirected edge between the two nodes $x$ and $y$ . These edges form a tree of $n$ nodes.
The next line contains $n$ space-separated integers denoting the array $a$ $(0 \le a_i \le 10^9)$ .
输出格式
Output $n$ integers, where the $i$ -th integer is $1$ if Alice wins the game when the tree is rooted at node $i$ , or $0$ otherwise.
输入输出样例
输入 #1
5 1 1 2 1 3 5 2 4 3 0 3 2 4 4
输出 #1
1 0 0 1 1
说明/提示
Let us calculate the answer for sample input with root node as 1 and as 2.
Root node 1
Alice always wins in this case. One possible gameplay between Alice and Bob is:
- Alice moves one present from node 4 to node 3.
- Bob moves four presents from node 5 to node 2.
- Alice moves four presents from node 2 to node 1.
- Bob moves three presents from node 2 to node 1.
- Alice moves three presents from node 3 to node 1.
- Bob moves three presents from node 4 to node 3.
- Alice moves three presents from node 3 to node 1.
Bob is now unable to make a move and hence loses.
Root node 2
Bob always wins in this case. One such gameplay is:
- Alice moves four presents from node 4 to node 3.
- Bob moves four presents from node 5 to node 2.
- Alice moves six presents from node 3 to node 1.
- Bob moves six presents from node 1 to node 2.
Alice is now unable to make a move and hence loses.
Root node 1
Alice always wins in this case. One possible gameplay between Alice and Bob is:
- Alice moves one present from node 4 to node 3.
- Bob moves four presents from node 5 to node 2.
- Alice moves four presents from node 2 to node 1.
- Bob moves three presents from node 2 to node 1.
- Alice moves three presents from node 3 to node 1.
- Bob moves three presents from node 4 to node 3.
- Alice moves three presents from node 3 to node 1.
Bob is now unable to make a move and hence loses.
Root node 2
Bob always wins in this case. One such gameplay is:
- Alice moves four presents from node 4 to node 3.
- Bob moves four presents from node 5 to node 2.
- Alice moves six presents from node 3 to node 1.
- Bob moves six presents from node 1 to node 2.
Alice is now unable to make a move and hence loses.