【LG4169】[Violet]天使玩偶/SJY摆棋子

题面

洛谷

题解

至于\(cdq\)分治的解法，以前写过

\(kdTree\)的解法好像还\(sb\)一些

就是记一下子树的横、纵坐标最值然后求一下点到矩形得到距离

之后再剪枝即可

为什么不吸氧还是跑不过啊

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <algorithm>

using namespace std;

inline int gi() {

    register int data = 0, w = 1;

    register char ch = 0;

    while (!isdigit(ch) && ch != '-') ch = getchar();

    if (ch == '-') w = -1, ch = getchar();

    while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();

    return w * data;

}

void chkmin(int &x, int y) { if (x > y) x = y; }

void chkmax(int &x, int y) { if (x < y) x = y; }

const int MAX_N = 1e6 + 5;

const double alpha = 0.75;

const int INF = 1e9;

struct Point { int x[2]; } p[MAX_N];

struct Node {

    int mn[2], mx[2], ls, rs, size;

    Point tp;

} t[MAX_N];

int N, M, rt, cur, top, WD, ans, rub[MAX_N];

bool operator < (const Point &l, const Point &r) { return l.x[WD] < r.x[WD]; }

int newnode() {

    if (top) return rub[top--];

    else return ++cur;

}

void pushup(int o) {

    int ls = t[o].ls, rs = t[o].rs;

    for (int i = 0; i <= 1; i++) {

        t[o].mn[i] = t[o].mx[i] = t[o].tp.x[i];

        if (ls) chkmin(t[o].mn[i], t[ls].mn[i]), chkmax(t[o].mx[i], t[ls].mx[i]);

        if (rs) chkmin(t[o].mn[i], t[rs].mn[i]), chkmax(t[o].mx[i], t[rs].mx[i]);

    }

    t[o].size = t[ls].size + t[rs].size + 1;

}

int build(int l, int r, int wd) {

    if (l > r) return 0;

    int o = newnode(), mid = (l + r) >> 1;

    WD = wd, nth_element(&p[l], &p[mid], &p[r + 1]), t[o].tp = p[mid];

    t[o].ls = build(l, mid - 1, wd ^ 1), t[o].rs = build(mid + 1, r, wd ^ 1);

    return pushup(o), o;

}

void pia(int o, int num) {

    if (t[o].ls) pia(t[o].ls, num);

    p[num + t[t[o].ls].size + 1] = t[o].tp, rub[++top] = o;

    if (t[o].rs) pia(t[o].rs, num + t[t[o].ls].size + 1);

}

void check(int &o, int wd) {

    if (alpha * t[o].size < t[t[o].ls].size || alpha * t[o].size < t[t[o].rs].size)

        pia(o, 0), o = build(1, t[o].size, wd);

}

void insert(int &o, int wd, Point tmp) {

    if (!o) return (void)(o = newnode(), t[o].tp = tmp, t[o].ls = t[o].rs = 0, pushup(o));

    if (t[o].tp.x[wd] < tmp.x[wd]) insert(t[o].r

91s, wd ^ 1, tmp);

    else insert(t[o].ls, wd ^ 1, tmp);

    pushup(o), check(o, wd);

}

int getdis(int o, Point tmp) {

    int res = 0;

    for (int i = 0; i <= 1; i++) res += max(0, tmp.x[i] - t[o].mx[i]) + max(0, t[o].mn[i] - tmp.x[i]);

    return res;

}

int dist(Point a, Point b) { return abs(a.x[0] - b.x[0]) + abs(a.x[1] - b.x[1]); }

void query(int o, Point tmp) {

    ans = min(ans, dist(tmp, t[o].tp));

    int dl = INF, dr = INF;

    if (t[o].ls) dl = getdis(t[o].ls, tmp);

    if (t[o].rs) dr = getdis(t[o].rs, tmp);

    if (dl < dr) {

        if (dl < ans) query(t[o].ls, tmp);

        if (dr < ans) query(t[o].rs, tmp);

    } else {

        if (dr < ans) query(t[o].rs, tmp);

        if (dl < ans) query(t[o].ls, tmp);

    }

}

int main () {

    N = gi(), M = gi();

    for (int i = 1; i <= N; i++) p[i].x[0] = gi(), p[i].x[1] = gi();

    rt = build(1, N, 0);

    while (M--) {

        Point tmp;

        int op = gi(); tmp.x[0] = gi(), tmp.x[1] = gi();

        if (op == 1) insert(rt, 0, tmp);

        else ans = INF, query(rt, tmp), printf("%d\n", ans);

    }

    return 0;

}