Prior to the BC has done a similar ladder game topic, the problem is moving to the root, the problem is moving to the leaves. In the same order, as long as the final state of the different parity of those positions to do NIM can. So the question gives a condition: the parity of all leaf depths is the same. It is also important to note that the last BC, the root node is not movable, so the root node is an end state node, and here the leaves can also be manipulated (can eat), then the leaf node can also continue to move downward, so they are not the final state node, That is to say this problem as long as the leaf node with the odd and even do nim can.
Therefore, if Nim and is already 0, has been able to meet the initiator will lose, and the topic said must be exchanged, so long as the Jenchie even the same node to do the exchange, statistics on the number of odd and even nodes C (CNT, 2) on the finish; The hand must be different from the leaves and the odd couple to change the past so that Nim and 0, using the nature of the XOR and map can be done. See the code for details:
1#include <stdio.h>2#include <algorithm>3#include <string.h>4#include <map>5#include <vector>6#include <iostream>7 using namespacestd;8 Const intN = 1e5 +5;9typedefLong Longll;Ten One intA[n]; Avector<int>G[n]; - intN; - intJi =-1; the voidDfsintUintFaintDeep ) - { - if(Ji! =-1)return ; - intFlag =0; + for(intI=0; I<g[u].size (); i++) - { + intv =G[u][i]; A if(V! =FA) at { -Flag =1; -DFS (V, u, deep +1); - } - } - if(Flag = =0) in { -Ji = deep%2; to } + } -vector<int>yes, no; the voidDFS2 (intUintFaintDeep ) * { $ if(Deep%2==ji) yes.push_back (A[u]);Panax Notoginseng ElseNo.push_back (A[u]); - for(intI=0; I<g[u].size (); i++) the { + intv =G[u][i]; A if(V! =FA) the { +DFS2 (V, u, deep +1); - } $ } $ } -ll Comb (intx) {if(X <2)return 0;return(ll) x* (x1) /2;} - the intMain () - {Wuyiscanf"%d",&n); the for(intI=1; i<=n;i++) scanf ("%d", A +i); - for(intI=2; i<=n;i++) Wu { - intv; Aboutscanf"%d",&v); $ G[i].push_back (v); - G[v].push_back (i); - } -Dfs1, -1,1); ADFS2 (1, -1,1); + inttemp =0; the for(intI=0; I<yes.size (); i++) Temp ^=Yes[i]; - if(temp = =0) $ { thell ans = comb (yes.size ()) +Comb (no.size ()); themap<int,int>Inyes; the for(intI=0; I<yes.size (); i++) inyes[yes[i]]++; the for(intI=0; I<no.size (); i++) ans + =(LL) inyes[no[i]]; -cout << ans <<Endl; in } the Else the { Aboutll ans =0; themap<int,int>Inno; the for(intI=0; I<no.size (); i++) inno[no[i]]++; the for(intI=0; I<yes.size (); i++) + { -ll t2 = Temp ^(Yes[i]); theAns + =Inno[t2];Bayi } thecout << ans <<Endl; the } - return 0; -}
Codeforces 812E Sagheer and Apple tree--(ladder game)