zoukankan      html  css  js  c++  java
  • POJ-2318 TOYS 计算几何 判断点在线段的位置

    题目链接:https://cn.vjudge.net/problem/POJ-2318

    题意

    在一个矩形内,给出n-1条线段,把矩形分成n快四边形
    问某些点在那个四边形内

    思路

    二分+判断点与位置关系

    提交过程

    WA*n x1和x2,y1和y2在复制的时候没分清(哭
    WA 可能存在二分问题?
    AC

    代码

    #define PI 3.1415926
    #include <cmath>
    #include <cstdio>
    #include <vector>
    #include <algorithm>
    using namespace std;
    const double eps=1e-10;
    const int maxn=5e3+20;
    
    struct Point{
    	double x, y;
    
    	Point(int x=0, int y=0):x(x), y(y) {}
    	// no known conversion for argument 1 from 'Point' to 'Point&'
    	Point operator + (Point p){return Point(x+p.x, y+p.y);}
    	Point operator - (Point p){return Point(x-p.x, y-p.y);}
    	Point operator * (double k){return Point(k*x, k*y);}
    	Point operator / (double k){return Point(x/k, y/k);}
    	bool operator < (Point p) const{return (x==p.x)?(y<p.y):(x<p.x);}	// need eps?
    	bool operator == (const Point p) const{return fabs(x-p.x)<eps&&fabs(y-p.y)<eps;}
    	double norm(void){return x*x+y*y;}
    	double abs(void){return sqrt(norm());}
    	double dot(Point p){return x*p.x+y*p.y;}		// cos
    	double cross(Point p){return x*p.y-y*p.x;}		// sin
    };
    struct Segment{Point p1, p2;};
    struct Circle{Point o; double rad;};
    typedef Point Vector;
    typedef vector<Point> Polygon;
    typedef Segment Line;
    
    int ccw(Point p0, Point p1, Point p2){
    	Vector v1=p1-p0, v2=p2-p0;
    	if (v1.cross(v2)>eps) return 1;			// anti-clockwise
    	if (v1.cross(v2)<-eps) return -1;		// clockwise
    	if (v1.dot(v2)<0) return 2;
    	if (v1.norm()<v2.norm()) return -2;
    	return 0;
    }
    
    Point project(Segment s, Point p){
    	Vector base=s.p2-s.p1;
    	double k=(p-s.p1).cross(base)/base.norm();
    	return s.p1+base*k;
    }
    
    Point reflect(Segment s, Point &p){
    	return p+(project(s, p)-p)*2;
    }
    
    double lineDist(Line l, Point p){
    	return abs((l.p2-l.p1).cross(p-l.p1)/(l.p2-l.p1).abs());
    }
    
    double SegDist(Segment s, Point p){
    	if ((s.p2-s.p1).dot(p-s.p1)<0) return Point(p-s.p1).abs();
    	if ((s.p1-s.p2).dot(p-s.p2)<0) return Point(p-s.p2).abs();
    	return abs((s.p2-s.p1).cross(p-s.p1)/(s.p2-s.p1).abs());
    }
    
    bool intersect(Point p1, Point p2, Point p3, Point p4){
    	return ccw(p1, p2, p3)*ccw(p1, p2, p4)<=0 &&
    			ccw(p3, p4, p1)*ccw(p3, p4, p2)<=0;
    }
    
    Point getCrossPoint(Segment s1, Segment s2){
    	Vector base=s2.p2-s2.p1;
    	double d1=abs(base.cross(s1.p1-s2.p1));
    	double d2=abs(base.cross(s1.p2-s2.p1));
    	double t=d1/(d1+d2);
    	return s1.p1+(s1.p2-s1.p1)*t;
    }
    
    double area(Polygon poly){
    	double res=0; long long size=poly.size();
    	for (int i=0; i<poly.size(); i++)
    		res+=poly[i].cross(poly[(i+1)%size]);
    	return abs(res/2);
    }
    
    int contain(Polygon poly, Point p){
    	int n=poly.size();
    	bool flg=false;
    	for (int i=0; i<n; i++){
    		Point a=poly[i]-p, b=poly[(i+1)%n]-p;
    		if (ccw(poly[i], poly[(i+1)%n], p)==0) return 1;	// 1 means on the polygon.
    		if (a.y>b.y) swap(a, b);
    		if (a.y<0 && b.y>0 && a.cross(b)>0) flg=!flg;
    	}return flg?2:0;										// 2 fo inner, 0 for outer.
    }
    
    Polygon convexHull(Polygon poly){
    	if (poly.size()<3) return poly;
    	Polygon upper, lower;
    	sort(poly.begin(), poly.end());
    	upper.push_back(poly[0]); upper.push_back(poly[1]);
    	lower.push_back(poly[poly.size()-1]); lower.push_back(poly[poly.size()-2]);
    	for (int i=2; i<poly.size(); i++){
    		for (int n=upper.size()-1; n>=1 && ccw(upper[n-1], upper[n], poly[i])!=-1; n--)
    			upper.pop_back();
    		upper.push_back(poly[i]);
    	}
    	for (int i=poly.size()-3; i>=0; i--){
    		for (int n=lower.size()-1; n>=1 && ccw(lower[n-1], lower[n], poly[i])!=-1; n--)
    			lower.pop_back();
    		lower.push_back(poly[i]);
    	}
    	for (int i=1; i<lower.size(); i++)
    		upper.push_back(lower[i]);
    	return upper;
    }
    
    Segment seg[maxn];
    int n, m;
    int solve(Point p){
    	int l=0, r=n;
    	while (l<r){
    		int mid=l+(r-l)/2;
    		if (ccw(seg[mid].p1, seg[mid].p2, p)==-1) r=mid;
    		else l=mid+1;
    	}
    	for (int i=max(l-3, 0); i<=min(l+3, n); i++)
    		if (ccw(seg[i].p1, seg[i].p2, p)==-1)
    			return l;
    }
    
    int main(void){
    	long long x, y, x1, y1, x2, y2, xt1, xt2;
    
    	while (scanf("%d", &n)==1 && n){
    		int bin[maxn]={0};
    		scanf("%d%lld%lld%lld%lld", &m, &x1, &y1, &x2, &y2);
    		for (int i=0; i<n; i++){
    			scanf("%lld%lld", &xt1, &xt2);
    			seg[i].p1=Point(xt1-x1, y1-y2);
    			seg[i].p2=Point(xt2-x1, y2-y2);
    		}
    		seg[n].p1=Point(x2-x1, y1-y2);
    		seg[n].p2=Point(x2-x1, y2-y2);
    
    		while (m--){
    			scanf("%lld%lld", &x, &y);
    			bin[solve(Point(x-x1, y-y2))]++;
    		}
    		for (int i=0; i<=n; i++)
    			printf("%d: %d
    ", i, bin[i]);
    		printf("
    ");
    	}
    
    	return 0;
    }
    
    Time Memory Length Lang Submitted
    204ms 716kB 4134 G++ 2018-08-01 12:19:23
  • 相关阅读:
    C# WCF的POST请求包含Stream及多个参数
    C# Fakes
    C# 计算文件的MD5
    C# 获取计算机的硬件、操作系统信息
    整数拆分
    整数拆分问题的四种解法【转载】
    python爬虫
    【转载】Scrapy安装及demo测试笔记
    Python野生库
    【转载】python3安装scrapy之windows32位爬坑
  • 原文地址:https://www.cnblogs.com/tanglizi/p/9403386.html
Copyright © 2011-2022 走看看