poj2826 An Easy Problem?!

题目描述：

vjudge

POJ

题解：

这道题告诉我们POJ的数据是极强的……

计算几何。

有好几个特殊情况，都在这组数据里面。

10
6259 2664 8292 9080 1244 2972 9097 9680

0 1 1 0
1 0 2 1

0 1 2 1
1 0 1 2

0 0 10 10
0 0 9 8

0 0 10 10
0 0 8 9

0.9 3.1 4 0
0 3 2 2

0 0 0 2
0 0 -3 2

1 1 1 4
0 0 2 3

1 2 1 4
0 0 2 3

0 0 1 1
0 0 1 2

感谢好人，好人

代码：

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const double eps = 1e-8;
int dcmp(double x)
{
    if(fabs(x)<=eps)return 0;
    return x>0?1:-1;
}
struct Point
{
    double x,y;
    Point(){}
    Point(double x,double y):x(x),y(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 * (const double&a)const{return Point(x*a,y*a);}
    double operator ^ (const Point&a)const{return x*a.y-y*a.x;}
}a,b,c,d;
typedef Point Vector;
struct Line
{
    Point p;
    Vector v;
    Line(){}
    Line(Point p,Vector v):p(p),v(v){}
}s,t;
int n;
bool diff(Line l,Point a,Point b)
{
    return dcmp(l.v^(a-l.p))*(l.v^(b-l.p))<=0;
}
Point L_L(Line a,Line b)
{
    double t = ((b.p-a.p)^b.v)/(a.v^b.v);
    return a.p+a.v*t;
}
Point L_Y(Line a,double y)
{
    return a.p+a.v*((y-a.p.y)/a.v.y);
}
void work()
{
    scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y,&d.x,&d.y);
    if(a.y<b.y)swap(a,b);
    if(c.y<d.y)swap(c,d);
    if(a.y<c.y)swap(a,c),swap(b,d);
    s = Line(a,b-a),t = Line(c,d-c);
    if(!dcmp(a.y-b.y)||!dcmp(c.y-d.y)||!diff(s,c,d)||!diff(t,a,b)||dcmp(a.x-c.x)*dcmp((a-b)^(c-d))<=0)
    {
        puts("0.00");
    }else
    {
        Point p = L_L(s,t),pp = L_Y(s,c.y);
        printf("%.2lf
",fabs(((c-p)^(pp-p))/2)+eps);
    }
}
int main()
{
    scanf("%d",&n);
    while(n--)work();
    return 0;
}