HDOJ 5996 dingyeye loves stone(階梯博弈)

dingyeye loves stone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 145    Accepted Submission(s): 87

Problem Description

dingyeye loves play stone game with you.

dingyeye has an n-point tree.The nodes are numbered from 0 to n−1,while the root is numbered 0.Initially,there are a[i] stones on the i-th node.The game is in turns.When one move,he can choose a node and move some(this number cannot be 0) of the stones on it to its father.One loses the game if he can't do anything when he moves.

You always move first.You want to know whether you can win the game if you play optimally.

 

Input

In the first line, there is an integer T indicating the number of test cases.

In each test case,the first line contains one integer n refers to the number of nodes.

The next line contains n−1 integers fa[1]⋯fa[n−1],which describe the father of nodes 1⋯n−1(node 0 is the root).It is guaranteed that 0≤fa[i]<i.

The next line contains n integers a[0]⋯a[n−1],which describe the initial stones on each nodes.It is guaranteed that 0≤a[i]<134217728.

1≤T≤100,1≤n≤100000.

It is guaranteed that there is at most 7 test cases such that n>100.

 

Output

For each test case output one line.If you can win the game,print "win".Ohterwise,print "lose".

 

Sample Input

2 2 0 1000 1 4 0 1 0 2 3 3 3

 

Sample Output

win lose

 

題意：請戳：中文題面 

題解：階梯博弈裸題啊， 可是沒學過，卒。。 請戳→ 階梯博弈   寫的挺詳細的一篇博文，有幾點筆誤。 但是很清楚的講了奇偶性對整體的影響及證明。。看看就懂了

代碼如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1e5+10; 
int d[maxn],fa[maxn];

int dfs(int x)//按照題意fa[a]<a，也可以不用這麼找深度 
{
	if(d[x]!=-1)
		return d[x];
	return d[x]=dfs(fa[x])+1;
}

int main()
{
	int t,n;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(int i=1;i<n;++i)
			scanf("%d",&fa[i]);
		memset(d,-1,sizeof(d));
		d[0]=0;
		for(int i=1;i<n;++i)
		{
			if(d[i]==-1)
				d[i]=dfs(i);
		}
		int ans=0,num;
		for(int i=0;i<n;++i)
		{
			scanf("%d",&num);
			if(d[i]&1)
				ans^=num;
		}
		if(ans)
			printf("win\n");
		else
			printf("lose\n");
	}
	return 0;
}