【LG4631】[APIO2018]Circle selection 選圓圈

題面

洛谷

題解

用\(kdt\)亂搞剪枝。
維護每個圓在\(x、y\)軸的座標範圍
相當於維護一個矩形的座標範圍為\([x-r,x+r],[y-r,y+r]\)
可以減小搜尋範圍
然後再判斷一下一個圓是否在當前搜尋的矩形內，不在就剪枝

#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; }
#define sqr(x) (1ll * (x) * (x)) 
const int MAX_N = 3e5 + 10;
struct Point { int x[2], r, id; } p[MAX_N];
struct Node {
    int mn[2], mx[2], ls, rs;
    Point tp; 
} t[MAX_N];
int N, WD, rt, ans[MAX_N], rub[MAX_N], cur, top;
inline bool operator < (const Point &l, const Point &r) { return l.x[WD] < r.x[WD]; }
inline bool cmp(const Point &l, const Point &r) { return (l.r == r.r) ? (l.id < r.id) : (l.r > r.r); }
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++) {
        int r = t[o].tp.r; 
        t[o].mx[i] = t[o].tp.x[i] + r; t[o].mn[i] = t[o].tp.x[i] - r; 
        if (ls) chkmax(t[o].mx[i], t[ls].mx[i]), chkmin(t[o].mn[i], t[ls].mn[i]); 
        if (rs) chkmax(t[o].mx[i], t[rs].mx[i]), chkmin(t[o].mn[i], t[rs].mn[i]); 
    } 
}
int build(int l, int r, int wd) {
    if (l > r) return 0;
    int mid = (l + r) >> 1, o = newnode(); 
    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; 
}
bool chk1(int o, Point tmp) { return sqr(tmp.x[0] - t[o].tp.x[0]) + sqr(tmp.x[1] - t[o].tp.x[1]) <= sqr(tmp.r + t[o].tp.r); } 
bool chk2(int o, Point tmp) {
    int x0 = tmp.x[0], x1 = tmp.x[1], r = tmp.r; 
    if (x0 + r < t[o].mn[0]) return 1;
    if (x0 - r > t[o].mx[0]) return 1; 
    if (x1 + r < t[o].mn[1]) return 1;
    if (x1 - r > t[o].mx[1]) return 1;
    return 0; 
} 
void query(int o, Point tmp) { 
    if (chk2(o, tmp)) return ; 
    if (!ans[t[o].tp.id] && chk1(o, tmp)) ans[t[o].tp.id] = tmp.id;
    if (t[o].ls) query(t[o].ls, tmp);
    if (t[o].rs) query(t[o].rs, tmp); 
}

int main () {
    N = gi();
    for (int i = 1; i <= N; i++) p[i].x[0] = gi(), p[i].x[1] = gi(), p[i].r = gi(), p[i].id = i; 
    rt = build(1, N, 0); 
    sort(p + 1, p + N + 1, cmp);
    for (int i = 1; i <= N; i++) if (!ans[p[i].id]) query(rt, p[i]);
    for (int i = 1; i <= N; i++) printf("%d ", ans[i]);
    printf("\n"); 
    return 0; 
}