HDU 5996 dingyeye loves stone （树形博弈）

dingyeye loves stone

                                                                 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                  Total Submission(s): 59    Accepted Submission(s): 34

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 be0) 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(node0 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

2201000 140 1 02 3 3 3

 

Sample Output

winlose传送门： http://acm.hdu.edu.cn/showproblem.php?pid=5996

思路： 树形博弈。 对于第 2、4、6层的数不用考虑（连续上移两次即可）；对于第1、3、5、7层的数，如果全部节点权值抑或，最后得到0（对一个高度为奇数的点进行的操作，即移动到高度为偶数进行处理）。

#include <queue>#include <functional>#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <stack>#include <vector>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>#include <ctime>#include <assert.h>#define REP(i,k,n) for(int i=k;i<n;i++)#define REPP(i,k,n) for(int i=k;i<=n;i++)#define scan(d) scanf("%d",&d)#define scann(n,m) scanf("%d%d",&n,&m)#define mst(a,k)  memset(a,k,sizeof(a));#define LL long long#define eps 1e-8#define INF 0x3f3f3f3f#define inf INF*INF#define upmo(a,b) (((a)=((a)+(b))%mo)<0?(a)+=mo:(a))#define mod 1000000007using namespace std;#define N 100005int f[N];int num[N];int ans;int main(){  int t;  scan(t);  while(t--){   int n;   scan(n);   REP(i,1,n){   int nn;   scan(nn);   f[i] = f[nn] + 1;   }   REP(i,0,n) scan(num[i]);   ans = 0;   REP(i,1,n){   if(f[i]&1) ans^=num[i];   }   if(!ans) printf("lose\n");   else printf("win\n");  }}