题解
不旋转坐标系，TLE，旋转坐标系，最慢一个点0.5s……maya，出题人数据水平很高了……
好吧，如果你不旋转坐标系，写一个正确性和复杂度未知的K - D树，没有优化，你可以得到87分的好成绩
但是你就是傻逼，你就是写不出来，能有什么办法，APIO Ag滚粗了呗= =
这道题看起来需要用什么东西维护一下平面，查找给定一个圆这个平面内多少个圆和它有交集，可以K - D树
我们考虑维护一个集合里的圆覆盖的矩形，就是最大的横纵坐标和最小的横纵坐标，查询的时候只要看看和当前圆横纵坐标是不是有交集，有交集就查，同时维护一下这棵树里还有没有圆没有被删除
当然复杂度可能会很劣……但是不劣的算法也不会，你就写了，写完之后87
哦87挺好的哦，然而不是考场上我想写写100分啊，看了大家似乎旋转了坐标系，然后我也转了，咦我怎么WA了……把eps调小一点就过了
为什么我那么菜啊= =
代码
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <map>
//#define ivorysi
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define mo 974711
#define MAXN 300005
#define eps 1e-3
#define RG register
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {putchar('-');x = -x;}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
struct Point {
db x,y;
Point() {}
Point(db _x,db _y) {
x = _x;y = _y;
}
};
struct Circle {
Point a;db R;int id;
}cir[MAXN],C[MAXN];
db o(db x) {
return x * x;
}
db dist(Point a,Point b) {
return o(a.x - b.x) + o(a.y - b.y);
}
bool cmpc(Circle s,Circle t) {
return s.R > t.R || (s.R == t.R && s.id < t.id);
}
const db alpha = acos(-1.0) / 5;
int tr[MAXN * 4],pos[MAXN],N,D[MAXN],tot,a[MAXN];
bool vis[MAXN],dimension;
struct node {
Circle cir;Point r1,r2;
node *lc,*rc;int siz;
void update_siz() {
siz = 0;
if(!vis[cir.id]) siz = 1;
siz += lc->siz;
siz += rc->siz;
}
void update() {
r1.x = cir.a.x - cir.R;r1.y = cir.a.y - cir.R;
r2.x = cir.a.x + cir.R;r2.y = cir.a.y + cir.R;
r1.x = min(r1.x,min(lc->r1.x,rc->r1.x));
r1.y = min(r1.y,min(lc->r1.y,rc->r1.y));
r2.x = max(r2.x,max(lc->r2.x,rc->r2.x));
r2.y = max(r2.y,max(lc->r2.y,rc->r2.y));
}
}*rt,*null;
bool cmp(Circle s,Circle t) {
if(dimension) return s.a.x + s.R + eps < t.a.x + t.R;
else return s.a.y + s.R + eps < t.a.y + t.R;
}
node* build(int l,int r,bool d) {
if(l > r) return null;
node *res = new node;
int mid = (l + r) >> 1;
dimension = d;
nth_element(C + l,C + mid,C + r + 1,cmp);
res->cir = C[mid];
res->lc = build(l,mid - 1,d ^ 1);res->rc = build(mid + 1,r,d ^ 1);
res->update();res->update_siz();
return res;
}
bool fit(node *u,Circle S) {
db l[2] = {u->r1.x,S.a.x - S.R};
db r[2] = {u->r2.x,S.a.x + S.R};
if(l[0] > l[1]) swap(l[0],l[1]),swap(r[0],r[1]);
if(l[1] > r[0]) return 0;
l[0] = u->r1.y,l[1] = S.a.y - S.R;
r[0] = u->r2.y,r[1] = S.a.y + S.R;
if(l[0] > l[1]) swap(l[0],l[1]),swap(r[0],r[1]);
if(l[1] > r[0]) return 0;
return 1;
}
void Query(node *u,Circle S) {
if(!u->siz) return;
if(!vis[u->cir.id]) {
if(dist(u->cir.a,S.a) <= o(u->cir.R + S.R) + eps) {
vis[u->cir.id] = 1;
a[u->cir.id] = S.id;
--tot;
}
}
if(fit(u->lc,S)) Query(u->lc,S);
if(fit(u->rc,S)) Query(u->rc,S);
u->update_siz();
}
void Init() {
null = new node;
null->siz = 0;
null->r1 = Point(2000000000,2000000000);
null->r2 = Point(-2000000000,-2000000000);
read(N);
db x,y,r;
for(int i = 1 ; i <= N ; ++i) {
scanf("%lf%lf%lf",&x,&y,&r);
cir[i] = (Circle){Point(x * cos(alpha) - y * sin(alpha),x * sin(alpha) + y * cos(alpha)),r,i};
C[i] = cir[i];
}
sort(cir + 1,cir + N + 1,cmpc);
rt = build(1,N,0);
}
void Solve() {
for(int i = 1 ; i <= N ; ++i) {
if(!a[cir[i].id]) Query(rt,cir[i]);
}
for(int i = 1 ; i <= N ; ++i) {
out(a[i]);if(i == N) enter;else space;
}
//out(clock());enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Init();
Solve();
return 0;
}