POJ - 2826 An Easy Problem?! （線段與線段相交求交點模板+分類討論）

鏈接：https://cn.vjudge.net/problem/POJ-2826

題意：給你兩個線段，問能接住多少從y軸正方向倒下來的水？

思路：線段與線段相交求交點。具體可以看kaungbin大神的題解。https://www.cnblogs.com/kuangbin/p/3192511.html

#include <cstdio>
#include <iostream>
#include <cmath>
#define ll long long
using namespace std;
const double eps = 1e-8;
const double N = 10005;
int sgn(double x)
{
	if(fabs(x)<eps) return 0;
	else if(x<0) return -1;
	else return 1;
}

struct Point
{
	double x,y;
	Point(){}
	Point(double x,double y):x(x),y(y){}
	Point operator -(const Point& b)const
	{
		return Point(x-b.x,y-b.y);
	}
	double operator ^(const Point& b)const
	{
		return x*b.y-y*b.x;
	}
	double operator *(const Point& b)const
	{
		return x*b.x+y*b.y;
	}
}a,b,c;
struct Line
{
	Point s,e;
	Line(){}
	Line(Point ss,Point ee)
	{
		s=ss,e=ee;
	}
	//求直線或線段（已確定相交）的交點 
	pair<Point,int> operator &(const Line& b)const
	{
		Point res=s;
		if(sgn((s-e)^(b.s-b.e))==0)
		{
			if(sgn((b.s-s)^(b.e-s))==0)//兩直線重合
				return make_pair(res,0);
			else return make_pair(res,1);//兩直線平行
		}
		double t=((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));
		res.x+=(e.x-s.x)*t;
		res.y+=(e.y-s.y)*t;
		return make_pair(res,2);//兩直線交點
	}
}l1,l2,l;
int n;
pair<Point,int> temp;
bool inter(Line l1,Line l2)
{
	return max(l1.s.x,l1.e.x)>=min(l2.s.x,l2.e.x)
	&& max(l2.s.x,l2.e.x)>=min(l1.s.x,l1.e.x)
	&& max(l1.s.y,l1.e.y)>=min(l2.s.y,l2.e.y)
	&& max(l2.s.y,l2.e.y)>=min(l1.s.y,l1.e.y)
	&& sgn((l2.s-l1.s)^(l1.e-l1.s))*sgn((l2.e-l1.s)^(l1.e-l1.s))<=0
	&& sgn((l1.s-l2.s)^(l2.e-l2.s))*sgn((l1.e-l2.s)^(l2.e-l2.s))<=0;
}
int main(void)
{
	int t;
	double x1,x2,x3,x4,y1,y2,y3,y4;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4);
		l1=Line(Point(x1,y1),Point(x2,y2));
		l2=Line(Point(x3,y3),Point(x4,y4));
		if(sgn(l1.s.y-l1.e.y)==0||sgn(l2.s.y-l2.e.y)==0)
		{
			puts("0.00");
			continue;
		}
		if(!inter(l1,l2))
		{
			puts("0.00");
			continue;
		}		
		if(l1.s.y<l1.e.y) swap(l1.s,l1.e);
		if(l2.s.y<l2.e.y) swap(l2.s,l2.e);	
			
		l=Line(l1.s,Point(l1.s.x,N));
		if(inter(l,l2))
		{
			puts("0.00");
			continue;			
		} 
		
		l=Line(l2.s,Point(l2.s.x,N));
		if(inter(l,l1))
		{
			puts("0.00");
			continue;			
		} 
		
		a=(l1&l2).first;
		if(sgn(l1.s.y-l2.s.y)<=0)
		{
			b=l1.s;
			l=Line(l1.s,Point(-N,l1.s.y));
			temp=l&l2;
			if(temp.second==2)
				c=temp.first;
			else
			{ 
				l=Line(l1.s,Point(N,l1.s.y));	
				c=(l&l2).first; 
			} 
		}
		else
		{
			b=l2.s;
			l=Line(l2.s,Point(-N,l2.s.y));
			temp=l&l1;
			if(temp.second==2)
				c=temp.first;
			else
			{ 
				l=Line(l2.s,Point(N,l2.s.y));	
				c=(l&l1).first; 
			} 				
		}
		printf("%.2f\n",fabs((b-a)^(c-a))/2.0+eps);
	}
	return 0;
}