CF893F Subtree Minimum Query 主席树

如果是求和就很好做了...

不是求和也无伤大雅....

一维太难限制条件了，考虑二维限制

一维$dfs$序，一维$dep$序

询问$(x, k)$对应着在$dfs$上查$[dfn[x], dfn[x] + sz[x] - 1]$，在$dep$序上查$[dep[x], dep[x] + k]$

这样子，每个询问对应查询一段矩形内的最小值

然而树套树是过不了的.....

发现一个询问看似在$dep$序上对应了一段区间，实际上可以扩展到对应一段前缀

这样子，只需要一个主席树就可以做到了

复杂度$O(n \log n)$

#include <map>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
namespace remoon {
    #define ri register int
    #define tpr template <typename ra>
    #define rep(iu, st, ed) for(ri iu = st; iu <= ed; iu ++)
    #define drep(iu, ed, st) for(ri iu = ed; iu >= st; iu --)    
    #define gc getchar
    inline int read() {
        int p = 0, w = 1; char c = gc();
        while(c > '9' || c < '0') { if(c == '-') w = -1; c = gc(); }
        while(c >= '0' && c <= '9') p = p * 10 + c - '0', c = gc();
        return p * w;
    }
    int wr[50], rw;
    #define pc(iw) putchar(iw)
    tpr inline void write(ra o, char c = '\n') {
        if(!o) pc('0');
        if(o < 0) o = -o, pc('-');
        while(o) wr[++ rw] = o % 10, o /= 10;
        while(rw) pc(wr[rw --] + '0');
        pc(c);
    }
    tpr inline void cmin(ra &a, ra b) { if(a > b) a = b; }
    tpr inline void cmax(ra &a, ra b) { if(a < b) a = b; } 
    tpr inline bool ckmin(ra &a, ra b) { return (a > b) ? a = b, 1 : 0; }
    tpr inline bool ckmax(ra &a, ra b) { return (a < b) ? a = b, 1 : 0; }
}
using namespace std;
using namespace remoon;

#define sid 300050
#define oid 12050000

int dfn[sid], sz[sid];
int n, r, m, id, tim, cnp, mxd;
int rt[sid], nxt[sid], node[sid], cap[sid];
int fa[sid], q[sid], w[sid], dep[sid];
int ls[oid], rs[oid], miv[oid];

inline void addedge(int u, int v) {
    nxt[++ cnp] = cap[u]; cap[u] = cnp; node[cnp] = v;
}

#define cur node[i]
inline void dfs(int o, int f) {
    fa[o] = f; dep[o] = dep[f] + 1;
    sz[o] = 1; dfn[o] = ++ tim;
    for(int i = cap[o]; i; i = nxt[i])
    if(cur != f) dfs(cur, o), sz[o] += sz[cur];
}

inline void insert(int &now, int pre, int l, int r, int p, int v) {
    now = ++ id;
    ls[now] = ls[pre]; rs[now] = rs[pre];
    miv[now] = min(miv[pre], v);
    if(l == r) return;
    int mid = (l + r) >> 1;
    if(p <= mid) insert(ls[now], ls[pre], l, mid, p, v);
    else insert(rs[now], rs[pre], mid + 1, r, p, v);
}

inline void build() {    
    int fr = 1, to = 0; 
    q[++ to] = r; miv[0] = 1e9;
    while(fr <= to) {
        int o = q[fr];
        for(ri i = cap[o]; i; i = nxt[i])
        if(cur != fa[o]) q[++ to] = cur;
        if(dep[o] != dep[q[fr - 1]]) 
        insert(rt[dep[o]], rt[dep[o] - 1], 1, n, dfn[o], w[o]);
        else insert(rt[dep[o]], rt[dep[o]], 1, n, dfn[o], w[o]);
        fr ++; cmax(mxd, dep[o]);
    }
}

inline int qry(int o, int l, int r, int ml, int mr) {
    if(ml > r || mr < l || !o) return 1e9;
    if(ml <= l && mr >= r) return miv[o];
    int mid = (l + r) >> 1;
    return min(qry(ls[o], l, mid, ml, mr), qry(rs[o], mid + 1, r, ml, mr));
}

int main() {
    n = read(); r = read();
    rep(i, 1, n) w[i] = read();
    rep(i, 2, n) {
        int u = read(), v = read();
        addedge(u, v); addedge(v, u);
    }
    dfs(r, 0); build();
    int lst = 0; m = read();
    rep(i, 1, m) {
        int x = (read() + lst) % n + 1;
        int k = (read() + lst) % n;
        write(lst = qry(rt[min(dep[x] + k, mxd)], 1, n, dfn[x], dfn[x] + sz[x] - 1));
    }
    return 0;
}