「vijos-bashu」lxhgww的奇思妙想(长链剖分)

倍增离线，预处理出爹和孙子们。查询\(O(1)\)

#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <algorithm>
#include <iostream>
#define R(a,b,c) for(register int a = (b); a <= (c); ++a)
#define nR(a,b,c) for(register int a = (b); a >= (c); --a)
#define Swap(a,b) ((a) ^= (b) ^= (a) ^= (b))
#define MP make_pair
#ifdef QWQ
#define FileOpen() freopen("in.txt", "r", stdin)
#define FileSave() freopen("out.txt", "w", stdout)
#define D_e_Line cerr << "\n--------\n"
#define D_e(x) cerr << (#x) << " : " << x << endl
#define C_e(x) cout << (#x) << " : " << x << endl
#define Pause() system("pause")
#define TIME() fprint(stderr, "TIME : %.3lfms\n", (double)clock() / (double)CLOCKS_PER_SEC)
#include <cassert>
#else
#define FileOpen()
#define FileSave()
#define D_e_Line
#define D_e(x)
#define C_e(x)
#define Pause()
#define TIME()
#endif
struct FastIO {
    template<typename ATP> inline FastIO & operator >> (ATP & x)  {
        x = 0; int f = 1; char c;
        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-') f = -1;
        while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
        if(f == -1) x = -x;
        return *this;
    }
} io;
using namespace std;
template<typename ATP> inline ATP Max(ATP x, ATP y) {
    return x > y ? x : y;
}
template<typename ATP> inline ATP Min(ATP x, ATP y) {
    return x < y ? x : y;
}
template<typename ATP> inline ATP Abs(ATP x) {
    return x > 0 ? x : -x;
}
#include <vector>
const int N = 3e5 + 7;
vector<int> D[N], U[N];
struct Edge {
    int nxt, pre;
} e[N << 1];
int head[N], cntEdge;
inline void add(int u, int v) {
    e[++cntEdge] = (Edge){ head[u], v}, head[u] = cntEdge;
}
int n;
int f[N][21], dep[N], md[N], len[N], son[N], top[N];
inline void DFS_First(int u, int father) {
    f[u][0] = father, md[u] = dep[u] = dep[father] + 1;
    R(i,1,19){
        if(f[u][i - 1]) f[u][i] = f[f[u][i - 1]][i - 1];
        else break;
    }
    for(register int i = head[u]; i; i = e[i].nxt){
        int v = e[i].pre;
        if(v == father) continue;
        DFS_First(v, u);
        if(md[v] > md[son[u]]) son[u] = v, md[u] = md[v];
    }
}
void DFS_Second(int u, int Tp) {
    top[u] = Tp, len[u] = md[u] - dep[Tp] + 1;
    if(!son[u]) return;
    DFS_Second(son[u], Tp);
    for(register int i = head[u]; i; i = e[i].nxt){
        int v = e[i].pre;
        if(v != f[u][0] && v != son[u])
            DFS_Second(v, v);
    }
}
int H[N];
inline void Init() {
    int now = 0;
    R(i,1,n){
        if(!(i & (1 << now))) ++now;
        H[i] = now;
    }
    R(i,1,n){
        if(i == top[i]){
            for(register int j = 1, u = i; j <= len[i] && u; ++j) u = f[u][0], U[i].push_back(u);
            for(register int j = 1, u = i; j <= len[i] && u; ++j) u = son[u], D[i].push_back(u);
        }
    }
}
inline int Query(int u, int K) {
    if(K > dep[u]) return 0;
    if(!K) return u;
    u = f[u][H[K]], K ^= (1 << H[K]);
    if(!K) return u;
    if(dep[u] - dep[top[u]] == K) return top[u];
    if(dep[u] - dep[top[u]] < K) return U[top[u]][K - dep[u] + dep[top[u]] - 1];
    return D[top[u]][dep[u] - dep[top[u]] - K - 1];
}
int main() {
    io >> n;
    R(i,2,n){
        int u, v;
        io >> u >> v;
        add(u, v);
        add(v, u);
    }
    DFS_First(1, 0);
    DFS_Second(1, 1);
    Init();
    int lst = 0, Q;
    io >> Q;
    while(Q--){
        int u, K;
        io >> u >> K;
        u ^= lst, K ^= lst;
        printf("%d\n", lst = Query(u, K));
    }
    return 0;
}