BZOJ 2618 CQOI 2006 凸多边形 半平面交

标签：bzoj   cqoi2006   计算几何   半平面交   

题目大意：给出n个凸多边形，求这些多边形的面积的交。

思路：犯傻了。。以后看到凸多边形第一时间就要想到半平面交啊。。多明显啊，半天愣着没想出来。

CODE：

#include <cmath>
#include <cstdio>
#include <iomanip>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 6100
#define EPS 1e-10
#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);
	}
	void Read() {
		scanf("%lf%lf",&x,&y);
	}
}p[MAX],temp[600];

struct Line{
	Point p,v;
	double alpha;
	
	Line(Point _,Point __):p(_),v(__) {}
	Line() {}
	bool operator <(const Line &a)const {
		return alpha < a.alpha;
	}
	void Calc() {
		alpha = atan2(v.y,v.x);
	}
}line[MAX],q[MAX];

int k,cnt;
int lines;

inline double Cross(const Point &a,const Point &b)
{
	return a.x * b.y - a.y * b.x;
}

inline void MakeLine(const Point &a,const Point &b)
{
	line[++lines] = Line(a,b - a);
}

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 t = Cross(l2.v,u) / Cross(l1.v,l2.v);
	return l1.p + l1.v * t;
}

double HalfplanIntersection()
{
	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;
	p[tail] = GetIntersection(q[tail],q[front]);
	if(tail - front <= 1)	return .0;
	double re = .0;
	for(int i = front; i < tail; ++i)
		re += Cross(p[i],p[i + 1]);
	re += Cross(p[tail],p[front]);
	return re / 2.0;
}

int main()
{
	cin >> k;
	for(int i = 1; i <= k; ++i) {
		scanf("%d",&cnt);
		for(int j = 1; j <= cnt; ++j)
			temp[j].Read();
		for(int j = 1; j < cnt; ++j)
			MakeLine(temp[j],temp[j + 1]);
		MakeLine(temp[cnt],temp[1]);
	}
	for(int i = 1; i <= lines; ++i)
		line[i].Calc();
	sort(line + 1,line + lines + 1);
	cout << fixed << setprecision(3) << HalfplanIntersection() << endl;
	return 0;
}

BZOJ 2618 CQOI 2006 凸多边形 半平面交

标签：bzoj   cqoi2006   计算几何   半平面交   

原文地址：http://blog.csdn.net/jiangyuze831/article/details/41382251