Problem Description
Alice and Bob are playing a game called 'Gameia ? Gameia !'. The game goes like this :
- There is a tree with all node unpainted initial.
- Because Bob is the VIP player, so Bob has K chances to make a small change on the tree any time during the game if he wants, whether before or after Alice's action. These chances can be used together or separate, changes will happen in a flash. each change is defined as cut an edge on the tree.
- Then the game starts, Alice and Bob take turns to paint an unpainted node, Alice go first, and then Bob.
- In Alice's move, she can paint an unpainted node into white color.
- In Bob's move, he can paint an unpainted node into black color, and what's more, all the other nodes which connects with the node directly will be painted or repainted into black color too, even if they are white color before.
- When anybody can't make a move, the game stop, with all nodes painted of course. If they can find a node with white color, Alice win the game, otherwise Bob.
Given the tree initial, who will win the game if both players play optimally?
Input
The first line of the input gives the number of test cases T; T test cases follow.
Each case begins with one line with two integers N and K : the size of the tree and the max small changes that Bob can make.
The next line gives the information of the tree, nodes are marked from 1 to N, node 1 is the root, so the line contains N-1 numbers, the i-th of them give the farther node of the node i+1.
Limits
T≤100
1≤N≤500
0≤K≤500
1≤Pi≤i
Output
For each test case output one line denotes the answer.
If Alice can win, output "Alice" , otherwise "Bob".
Sample Input
2
2 1
1
3 1
1 2
Sample Output
Bob
Alice
题意：
Alice和Bob玩一个游戏，一开始有一颗有n个节点并且没有颜色的树，Bob和Alice分别对树上的节点进行染色，Alice每次将一个没有颜色的点涂成白色，Bob每次将一个没有颜色的点涂成黑色，并且可以将与涂上黑色的这个点直接相邻的点变为黑色，假如最后树被涂满之后还存在白色点，Alice赢，否则Bob赢。Bob还有一个特权，可以在任意时候，删除任意一条边，这样的特权可以使用k次。
分析：
首先当有一条长度大于等于3的链的时候，Alice一定可以获胜。可以找出来3 和 4 的时候这个结论是成立的。（3的时候先染中间那个点，4的时候先染最后一个点）当长度大于4的时候，Alice可以染倒数第二个点，那么Bob只能染最后一个点，否则他必输。这样就相当于将链的长度减小了2。那么接下去一定会变成3或4的链，Alice必胜。
实际上如果一个相连通的点集元素大于等于3的时候，Alice都是必胜的。因为如果是一棵树Alice可以用上述方式，每次染一个叶子节点的父节点，Bob只能染这个叶子结点。那么最后就会化成链或者只剩一个节点，那么她必胜。
对于一个节点连同它所有的子节点的个数如果是奇数的话，那么这样无论如何Alice都会赢，每次Alice先染白一个点，然后每次Bob涂的时候只有涂这个节点的子节点才能将这个节点再涂成黑色，这样他俩一轮下来只能涂两个，这样下来即使每个节点都是有两个子节点，最后被涂成白色的子节点个数最多为n/2-1个。
如果Bob的特权次数小于这个数的话，必然也是Alice赢。
如果为偶数的话，并且可以将这个树划分为所有都是只有两个点相连的话，Bob赢。
代码：
#include<stdio.h>
#include<vector>
using namespace std;
vector<int>vt[505];
int size[505];
int flag=0;
void dfs(int u)
{
int num=0;///奇数节点的个数
size[u]=1;
for(int i=0; i<vt[u].size(); i++)///递归所有的子节点
{
int to=vt[u][i];
dfs(to);
size[u]+=size[to];
if(size[to]%2==1)num++;///为奇数的节点（包括单独的叶子节点），Alice回赢
}
if(num>=2)flag=1;///这样的话也就相当于只有有两个节点的话，Bob才会赢
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
for(int i=0; i<505; i++)vt[i].clear();
int n,k;
scanf("%d%d",&n,&k);
for(int i=2; i<=n; i++)
{
int f;
scanf("%d",&f);
vt[f].push_back(i);
}
flag=0;
if(n%2==1||n/2-1>k)///奇数个或者特权数不够用，必定Alice赢
printf("Alice\n");
else///节点个数为偶数个并且特权个数足够用
{
dfs(1);
if(flag==1)
printf("Alice\n");
else
printf("Bob\n");
}
}
}