题意：给定一棵n个点的树，m次强制在线的询问，每次询问x的k级祖先的编号
n<=3e5，m<=1.8e6
思路：参考资料：https://zhuanlan.zhihu.com/p/25984772
https://blog.bill.moe/long-chain-subdivision-notes/
不知道为什么写成f[v][0]=u会RE一部分，f[u][0]=fa就AC了，下次写长链剖分的时候注意
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef unsigned int uint; 5 typedef unsigned long long ull; 6 typedef pair<int,int> PII; 7 typedef pair<ll,ll> Pll; 8 typedef vector<int> VI; 9 typedef vector<PII> VII; 10 typedef pair<ll,int>P; 11 #define N 300010 12 #define M 600010 13 #define fi first 14 #define se second 15 #define MP make_pair 16 #define pi acos(-1) 17 #define mem(a,b) memset(a,b,sizeof(a)) 18 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++) 19 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--) 20 #define lowbit(x) x&(-x) 21 #define Rand (rand()*(1<<16)+rand()) 22 #define id(x) ((x)<=B?(x):m-n/(x)+1) 23 #define ls p<<1 24 #define rs p<<1|1 25 26 const ll MOD=1e9+7,inv2=(MOD+1)/2; 27 double eps=1e-6; 28 ll INF=1ll<<62; 29 ll inf=5e13; 30 int dx[4]={-1,1,0,0}; 31 int dy[4]={0,0,-1,1}; 32 33 int f[N][20]; 34 vector<int> up[N],down[N]; 35 int head[M],vet[M],nxt[M],tot; 36 int dep[M],son[M],top[M],len[M],hb[M],mx[M]; 37 38 int read() 39 { 40 int v=0,f=1; 41 char c=getchar(); 42 while(c<48||57<c) {if(c=='-') f=-1; c=getchar();} 43 while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar(); 44 return v*f; 45 } 46 47 void add(int a,int b) 48 { 49 nxt[++tot]=head[a]; 50 vet[tot]=b; 51 head[a]=tot; 52 } 53 54 void dfs1(int u,int fa,int depth) 55 { 56 mx[u]=dep[u]=depth; 57 f[u][0]=fa; 58 rep(i,1,19) f[u][i]=f[f[u][i-1]][i-1]; 59 int e=head[u]; 60 while(e) 61 { 62 int v=vet[e]; 63 if(v!=fa) 64 { 65 dfs1(v,u,depth+1); 66 if(mx[v]>mx[son[u]]) 67 { 68 son[u]=v; 69 mx[u]=mx[v]; 70 } 71 } 72 e=nxt[e]; 73 } 74 } 75 76 void dfs2(int u,int ance,int l) 77 { 78 len[u]=l; 79 top[u]=ance; 80 if(son[u]) 81 { 82 dfs2(son[u],ance,l+1); 83 len[u]=len[son[u]]; 84 } 85 int e=head[u]; 86 while(e) 87 { 88 int v=vet[e]; 89 if(v!=f[u][0]&&v!=son[u]) dfs2(v,v,1); 90 e=nxt[e]; 91 } 92 } 93 94 int query(int u,int k) 95 { 96 if(k>dep[u]) return 0; 97 if(!k) return u; 98 u=f[u][hb[k]]; 99 k-=(1<<hb[k]); 100 if(!k) return u; 101 if(dep[u]-dep[top[u]]==k) return top[u]; 102 if(dep[u]-dep[top[u]]>k) return down[top[u]][dep[u]-dep[top[u]]-k-1]; 103 return up[top[u]][k-(dep[u]-dep[top[u]])-1]; 104 } 105 106 int main() 107 { 108 int n=read(); 109 rep(i,1,n) head[i]=0; 110 tot=0; 111 rep(i,1,n-1) 112 { 113 int x=read(),y=read(); 114 add(x,y); 115 add(y,x); 116 } 117 dep[1]=0; 118 dfs1(1,0,1); 119 dfs2(1,1,1); 120 rep(i,1,19) 121 rep(j,(1<<(i-1)),((1<<i)-1)) hb[j]=i-1; //highbit 122 rep(i,1,n) 123 if(top[i]==i) 124 { 125 int k=i; 126 rep(j,1,len[i]) 127 { 128 k=f[k][0]; 129 up[i].push_back(k); 130 } 131 k=i; 132 rep(j,1,len[i]) 133 { 134 k=son[k]; 135 down[i].push_back(k); 136 } 137 } 138 139 int m=read(),ans=0; 140 rep(i,1,m) 141 { 142 int x=read(),k=read(); 143 x^=ans; k^=ans; 144 ans=query(x,k); 145 printf("%d\n",ans); 146 } 147 return 0; 148 }