BZOJ2961 共点圆[CDQ分治]

题面

bzoj

其实就是推一下圆的式子
长成这个样子
假设要查询的点是(x, y) 某个圆心是(p, q)
\((x - p)^2 + (y - q)^2 \leq p^2 + q^2\)
变成
\(-\frac{2x}{2y}p + \frac{x^2+y^2}{2y} \leq q\)
那么一个点合法就要对所有圆心都满足上面这个式子
很明显拿斜率截就好啦
然后cdq维护上下凸包

附：cdq维护凸包过程

void cdq(int L, int R){
    if(L == R) return ;
    int Mid = L + ((R - L) >> 1), tl, tr;
    按照时间分左右
    cdq(L, Mid);
    int t1 = 0, t2 = 0;
    for(int i = L; i <= Mid; ++i){
	if(node[i].type) continue;
	构造上凸包 下凸包
    }
    for(int i = Mid + 1, j1 = 1, j2 = t2; i <= R; ++i){
	if(!node[i].type) continue;   
	    if(node[i].y > 0){//mdf
		根据斜率正负判断用上还是下（本题负斜率用下凸包
    }
    cdq(Mid + 1, R);	
    归并，按照x排序
}

如果你wa了你要知道
这道题不可以用叉积判凸包
这道题不可以做完减法判slope
这道题不可以不加eps
我哭辽qvq

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <cstring>
#define Sqr(x) ((x)*(x))
using namespace std;
const int N = 5e5 + 5;
const double eps = 1e-11;//这里1e-8到1e-11都可以
struct Node{
    int t;
	double x, y, k;	
	bool type, ok;
	friend bool operator <(Node x, Node y){return fabs(x.x - y.x) < eps ? x.y < y.y : x.x < y.x;}
}node[N], stk[N], s1[N], s2[N];
int n;
double slope(Node a,Node b)
{
    if(fabs(a.x-b.x)<eps)return a.y<b.y?1e18:-1e18;
    return (a.y-b.y)/(a.x-b.x);
}
double Dis(Node a,Node b){return sqrt(Sqr(a.x-b.x)+Sqr(a.y-b.y));}//距离和斜率必须这么写！不能减后判断！
//inline double cross(Node x, Node y){return x.x * y.y - x.y * y.x;}
bool rule_t(Node x, Node y){return x.t < y.t;}
bool rule_k(Node x, Node y){return x.k < y.k;}

void cdq(int L, int R){
    if(L == R) return ;
	int Mid = L + ((R - L) >> 1), tl, tr;
    tl = L, tr = Mid + 1;
    for(int i = L; i <= R; ++i)
		if(node[i].t <= Mid) stk[tl++] = node[i];
		else stk[tr++] = node[i]; 
    for(int i = L; i <= R; ++i) node[i] = stk[i];
    cdq(L, Mid);
    int t1 = 0, t2 = 0;
    for(int i = L; i <= Mid; ++i){
		if(node[i].type) continue;
	    while(t1 > 1 && slope(s1[t1 - 1], node[i]) - eps < slope(s1[t1 - 1], s1[t1])) --t1;
		s1[++t1] = node[i];//涓嬪嚫鍖咃紙璐熸枩鐜囷級	
	    while(t2 > 1 && slope(s2[t2 - 1], node[i]) + eps > slope(s2[t2 - 1], s2[t2])) --t2;//用叉积判凸包会wa！ 
		s2[++t2] = node[i];//涓婂嚫鍖?
	}
	for(int i = Mid + 1, j1 = 1, j2 = t2; i <= R; ++i){
	    if(!node[i].type) continue;   
		if(node[i].y > 0){//mdf
		    while(j1 < t1 && slope(s1[j1], s1[j1 + 1]) < node[i].k) ++j1;
		    if(j1 <= t1 && Dis(s1[j1], s1[0]) < Dis(node[i], s1[j1])) node[i].ok = 0;
		//		printf("-------\n");s1[j1].print(); node[i].print();
		} 	
		else {
		    while(j2 > 1 && slope(s2[j2], s2[j2 - 1]) > node[i].k) --j2;
			if(j2 >= 1 && Dis(s2[j2], s2[0]) < Dis(node[i], s2[j2]))	node[i].ok = 0;
		  //  printf("---------\n");s2[j2].print(); node[i].print();
		}
	}
    //printf("%d %d\n", L, R);
    cdq(Mid + 1, R);	
	tl = L, tr = Mid + 1, t1 = L;
	while(tl <= Mid && tr <= R)
		if(node[tl] < node[tr]) stk[t1++] = node[tl++];
		else stk[t1++] = node[tr++];
    while(tl <= Mid) stk[t1++] = node[tl++];
	while(tr <= R) stk[t1++] = node[tr++];
    for(int i = L; i <= R; ++i) node[i] = stk[i];
}

int main() {
	scanf("%d", &n);
	bool flag = 0;
	for(int i = 1; i <= n; ++i){
	    scanf("%d%lf%lf", &node[i].type, &node[i].x, &node[i].y);
		node[i].t = i;
		node[i].k = fabs(node[i].y) > eps ? -node[i].x / node[i].y : 
		(node[i].x > 0 ? -1e18 : 1e18);
		if(node[i].type) node[i].ok = flag; else flag = 1;
	}
	sort(node + 1, node + n + 1, rule_k);
	cdq(1, n);
	sort(node + 1, node + n + 1, rule_t);
	for(int i = 1; i <= n; ++i){
		//printf("%d %.4lf %.4lf %d\n", node[i].type, node[i].x, node[i].y, node[i].ok);
	    if(node[i].type){
		    if(node[i].ok) printf("Yes\n");
			else printf("No\n");	
		}	
	}
	system("PAUSE");
    return 0;
}