Codeforces 893F Subtree Minimum Query

炒鸡大水题，套路得要死（

看到「子树」，就有线段树合并和 DFS 序——考虑到维护的最小值没有区间减法，使用线段树合并。
强制在线，可持久化一下就好了。

代码：

#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 1e5;
int n,m,r;
int a[N + 5];
int to[(N << 1) + 5],pre[(N << 1) + 5],first[N + 5];
int lastans;
inline void add(int u,int v)
{
    static int tot = 0;
    to[++tot] = v,pre[tot] = first[u],first[u] = tot;
}
struct node
{
    int min;
    int ls,rs;
} seg[(N << 7) + 10];
int rt[N + 5];
int seg_tot;
void insert(int x,int k,int &p,int tl,int tr)
{
    if(!p)
        p = ++seg_tot,seg[p].min = 0x3f3f3f3f;
    seg[p].min = min(seg[p].min,k);
    if(tl == tr)
        return ;
    int mid = tl + tr >> 1;
    x <= mid ? insert(x,k,seg[p].ls,tl,mid) : insert(x,k,seg[p].rs,mid + 1,tr);
}
int query(int l,int r,int p,int tl,int tr)
{
    if(!p || (l <= tl && tr <= r))
        return seg[p].min;
    int mid = tl + tr >> 1;
    int ret = 0x3f3f3f3f;
    if(l <= mid)
        ret = min(ret,query(l,r,seg[p].ls,tl,mid));
    if(r > mid)
        ret = min(ret,query(l,r,seg[p].rs,mid + 1,tr));
    return ret;
}
int merge(int x,int y)
{
    if(!x || !y)
        return x | y;
    int p = ++seg_tot;
    seg[p].min = min(seg[x].min,seg[y].min);
    seg[p].ls = merge(seg[x].ls,seg[y].ls),seg[p].rs = merge(seg[x].rs,seg[y].rs);
    return p;
}
int fa[N + 5],dep[N + 5];
void dfs(int p)
{
    insert(dep[p],a[p],rt[p],1,n);
    for(register int i = first[p];i;i = pre[i])
        if(to[i] ^ fa[p])
            fa[to[i]] = p,dep[to[i]] = dep[p] + 1,dfs(to[i]),rt[p] = merge(rt[p],rt[to[i]]);
}
int main()
{
    seg[0].min = 0x3f3f3f3f;
    scanf("%d%d",&n,&r);
    for(register int i = 1;i <= n;++i)
        scanf("%d",a + i);
    int u,v;
    for(register int i = 1;i < n;++i)
        scanf("%d%d",&u,&v),add(u,v),add(v,u);
    dep[r] = 1,dfs(r);
    scanf("%d",&m);
    int x,k;
    while(m--)
    {
        scanf("%d%d",&x,&k);
        x = (x + lastans) % n + 1,k = (k + lastans) % n;
        printf("%d\n",lastans = query(dep[x],min(dep[x] + k,n),rt[x],1,n));
    }
}