HDU 6105 (2017 多校训练赛6 1010）Gameia（博弈）

2017 Multi-University Training Contest - Team 6 1010

Gameia

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

Alice and Bob are playing a game called 'Gameia ? Gameia !'. The game goes like this :
0. There is a tree with all node unpainted initial.
1. 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. 
2. Then the game starts, Alice and Bob take turns to paint an unpainted node, Alice go first, and then Bob.
3. In Alice's move, she can paint an unpainted node into white color.
4. 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.
5. 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

22 113 11 2

 

Sample Output

BobAlice

 

题意：
Alice和Bob玩游戏，有一个树，A可以选一个未染色结点染成白色  B可以在任何时候切断K条连边，B可以选一个未染色结点染成黑色同时这个结点连接的点也会变成黑色。最后还有白色的结点那么A赢否则B赢。分析：一开始觉得B很容易赢，但分析后发现，B其实只有在这颗树只有2个结点的时候才能赢否则必定输（不能切割的时候）因为刚开始的时候每个人的轮次是一样的，所以每次A走完B必定要覆盖A的点，那么只要有一回合A有空隙，或者A能分割B的决策那么A必赢。接着每切一刀相当将一颗树分成两颗，所以B要赢只有每颗数都是两个节点。那么直接从叶子结点找父亲结点切割，看需要多少刀就可以得到答案啦~

AC代码：

#include <cstdio>#include <queue>#include <cstring>#include <iostream>#include <string>#include <cstdlib>#include <algorithm>#include <vector>#include <map>#include <set>#include <ctime>#include <cmath>using namespace std;const int maxn=2000+5;int father[maxn];int vis[maxn];vector<int>g[maxn];int du[maxn],n;int toposort(){    memset(du,0,sizeof(du));    memset(vis,0,sizeof(vis));    int ans=0;    for(int i=1;i<=n;i++)    {    for(int j=0;j<g[i].size();j++)    {        du[i]++;    }    }    queue<int>Q;    for(int i=1;i<=n;i++)    {    if(!du[i])    {        Q.push(i);    }    }    while(!Q.empty())    {    int x=Q.front();    Q.pop();    int t=father[father[x]];//    printf("x=%d t=%d\n",x,t);    if(father[x]!=-1&&!vis[x]&&!vis[father[x]])    {        du[t]--;        vis[x]=1;        vis[father[x]]=1;        ans++;    }    else    {        return -1;    }    if(!du[t])    Q.push(t);        }    return ans;}int main(){    int T;    scanf("%d",&T);    while(T--)    {    for(int i=0;i<maxn;i++)        g[i].clear();    memset(father,-1,sizeof(father));    int k;    scanf("%d%d",&n,&k);    for(int i=2;i<=n;i++)    {        scanf("%d",&father[i]);        g[father[i]].push_back(i);    }    int ans=toposort();//    printf("!  %d\n",ans-1);    if(ans==-1||ans-1>k)        printf("Alice\n");    else        printf("Bob\n");    }}