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BZOJ 2732 HNOI 2012 射箭 半平面交

题目大意:给出一些与x轴垂直的线段,问一个经过原点的抛物线最多能按顺序经过多少条线段。


思路:总体上来说是数学题,我们来推一推。

设这个经过原点的抛物线为y = a * x ^ 2 + b * x,设一条线段的起点和终点为(x0,y1)和(x0,y2),且y2 > y1。

将x0带入到设出的抛物线中,会得到y = a * x0 ^ 2 + b * x0,这时候需要满足的是y <= y2 && y >= y1,也就是a * x0 ^ 2 + b * x0 <= y2 && y1 <= a * x0 ^ 2 + b * x0

整理一下思路,x0,y1,y2是已知量,a和b是我们设出来的量,不妨换一种写法,令x = a,y = b,k = x0,那么原不等式组就是

=> x * k ^ 2 + y * k <= y2 && x * k ^ 2 + y * k >= y1

=> x * k ^ 2 + y * k - y2 <= 0 && x * k ^2 + y * k - y1 >= 0

这样就很明显了,不等式组化成了两个半平面,之后利用半平面交判定是否存在就可以了。最外层套一个二分,时间复杂度大概是O(nlog^2n)

注意:此题卡精度,亲测1e-10会wa两个点,要1e-11以上才可以AC,同时一条直线的方向向量也要扩大n倍,否则就是被卡,别问我为什么,去给出题人寄刀片!


CODE:


#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 200010
#define EPS 1e-15
#define DCMP(a) (fabs(a) < EPS)
using namespace std;
 
struct Point{
    double x,y;
     
    Point(double _ = .0,double __ = .0):x(_),y(__) {}
    Point operator +(const Point &a)const {
        return Point(x + a.x,y + a.y);
    }
    Point operator -(const Point &a)const {
        return Point(x - a.x,y - a.y);
    }
    Point operator *(double a)const {
        return Point(x * a,y * a);
    }
}p[MAX];
 
struct Line{
    Point p,v;
    double alpha;
     
    Line(Point _,Point __):p(_),v(__) {
        alpha = atan2(v.y,v.x);
    }
    Line() {}
    bool operator <(const Line &a)const {
        return alpha < a.alpha;
    }
}src[MAX],line[MAX],q[MAX];
 
int asks;
int lines;
 
inline double Cross(const Point &p1,const Point &p2)
{
    return p1.x * p2.y - p1.y * p2.x;
}
 
inline bool OnLeft(const Point &p,const Line &l)
{
    return Cross(l.v,p - l.p) >= 0;
}
 
inline Point GetIntersection(const Line &l1,const Line &l2)
{
    Point u = l1.p - l2.p;
    double temp = Cross(l2.v,u) / Cross(l1.v,l2.v);
    return l1.p + l1.v * temp;
}
 
inline bool HalfplaneIntersection(int lines)
{
    int front = 1,tail = 1;
    q[1] = line[1];
    for(int i = 2; i <= lines; ++i) {
        while(front < tail && !OnLeft(p[tail - 1],line[i]))  --tail;
        while(front < tail && !OnLeft(p[front],line[i])) ++front;
        if(DCMP(Cross(q[tail].v,line[i].v)))
            q[tail] = OnLeft(q[tail].p,line[i]) ? q[tail]:line[i];
        else    q[++tail] = line[i];
        if(front < tail) p[tail - 1] = GetIntersection(q[tail],q[tail - 1]);
    }
    while(front < tail && !OnLeft(p[tail - 1],q[front])) --tail;
    return tail - front > 1;
}
 
inline bool Judge(int mid)
{
    mid <<= 1;
    memcpy(line + 1,src + 1,sizeof(Line) * mid);
    sort(line + 1,line + mid + 1);
    return HalfplaneIntersection(mid);
}
 
int main()
{
    cin >> asks;
    for(int i = 1; i <= asks; ++i) {
        static double x,y1,y2;
        scanf("%lf%lf%lf",&x,&y1,&y2);
        src[++lines] = Line(Point(0,y2 / x),Point(-1 / x,1));
        src[++lines] = Line(Point(0,y1 / x),Point(1 / x,-1));
    }
    int l = 1,r = asks,ans = 1;
    while(l <= r) {
        int mid = (l + r) >> 1;
        if(Judge(mid))
            ans = mid,l = mid + 1;
        else    r = mid - 1;
    }
    cout << ans << endl;
    return 0;
}


BZOJ 2732 HNOI 2012 射箭 半平面交