[BZOJ2716/2648][Violet 3]天使玩偶/SJY摆棋子[KDtree]

KDtree干这个复杂度是不对的，重构不一定有作用

解释一下的话，因为复杂度是跟size相关的，所以重构作用不大，KDtree在查询最近点对中的作用仅仅是剪枝，可以构造数据使得他遍历O(n)个节点

hack kdtree

（上面这个是按照洛谷数据范围 n,m 3e5造的

int n, m, x, y, now, ans, op, cnt, d[2];
struct Node {
  int Min[2], Max[2], d[2];
  Node *ls, *rs;
  Node (const pii &a, Node *b) {
    Min[0] = Max[0] = d[0] = a.fi;
    Max[1] = Min[1] = d[1] = a.se;
    ls = rs = b;
  }
  Node () {}
  inline void pushup() {
    lop0(i, 2) Min[i] = min(d[i], min(ls->Min[i], rs->Min[i])), Max[i] = max(d[i], max(ls->Max[i], rs->Max[i]));
  }
  inline bool operator < (const Node &rhs) const {
    return d[now] < rhs.d[now];
  }
} S[1300005], *null = S, *root, *root1;
inline void init() {
  null->Min[0] = null->Min[1] = 1e7, null->Max[0] = null->Max[1] = -1e7;//1e9的话 dis会爆int
  null->ls = null->rs = null;
  root1 = root = null;
}
inline Node *build(int l, int r) {
  if (l > r) return null;
  nth_element(S+l, S+mid, S+r+1);
  Node *cur = &S[mid];
  now ^= 1;
  cur->ls = build(l, mid-1), cur->rs = build(mid+1, r);
  return cur->pushup(), cur; 
}
inline void insert(Node *&cur, int *d, bool now) {
  if (cur == null) {
    cur = &(S[++cnt] = Node(mp(d[0],d[1]), null));
    return ;
  }
  insert(d[now] >= cur->d[now] ? cur->rs : cur->ls, d, !now);
  cur->pushup();
}
 
inline int dis(Node *cur, const pii &a) {
  return max(0, a.se - cur->Max[1]) + max(0, a.fi - cur->Max[0]) + max(0, cur->Min[0] - a.fi) + max(0, cur->Min[1] - a.se);
}
inline void query(Node *cur, const pii &a) {
  chmin(ans, abs(cur->d[0] - a.fi) + abs(cur->d[1] - a.se));
  int disl = dis(cur->ls, a), disr = dis(cur->rs, a);
  if (disl < disr) {
    if (disl < ans) query(cur->ls, a);
    if (disr < ans) query(cur->rs, a);
  }
  else {
    if (disr < ans) query(cur->rs, a);
    if (disl < ans) query(cur->ls, a);
  }
}
  
  
int main() {
  init();
  in, cnt, m;
  lop1(i, cnt) {
    in, x, y;
    S[i] = Node(mp(x, y), null);
  }
  root = build(1, cnt);
  while (m--) { 
    in, op, x, y;
    if (op == 1) {
      d[0] = x, d[1] = y;
      insert(root, d, 0);
    }
    else {
      ans = inft;
      query(root, mp(x, y));
      out, ans, '\n';
    }
  } 
  return 0;
}