[poj1410]Intersection

题目大意：求线段与实心矩形是否相交。

解题关键：转化为线段与线段相交的判断。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<iostream>
#define eps 1e-8
using namespace std;
typedef long long ll;
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;}
};
struct Line{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e){s=_s;e=_e;}
}A[];
int sgn(double x){
    if(fabs(x)<eps)return ;
    else if(x<) return -;
    else return ;
}
//判断线段相交，模板
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))<=&&
        sgn((l1.s-l2.s)^(l2.e-l2.s))*sgn((l1.e-l2.s)^(l2.e-l2.s))<=;
} 
 
int main(){
    int t,i;
    double xleft,ytop,xright,ybottom;
    double x1,y1,x2,y2;
    scanf("%d",&t);
    while(t--){
        scanf("%lf%lf%lf%lf",&A[].s.x,&A[].s.y,&A[].e.x,&A[].e.y);//线段
        scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
        xleft=min(x1,x2);xright=max(x1,x2);
        ybottom=min(y1,y2);ytop=max(y1,y2);
        A[].s.x=xleft;A[].s.y=ybottom;A[].e.x=xleft;A[].e.y=ytop;
        A[].s.x=xleft;A[].s.y=ytop;A[].e.x=xright;A[].e.y=ytop;
        A[].s.x=xright;A[].s.y=ytop;A[].e.x=xright;A[].e.y=ybottom;
        A[].s.x=xright;A[].s.y=ybottom;A[].e.x=xleft;A[].e.y=ybottom;//矩形的四条线段
        for(i=;i<=;++i) if(inter(A[],A[i]))break;
        bool flag=false;//矩形是实心的。
        if(A[].s.x<=xright&&A[].s.x>=xleft&&A[].s.y>=ybottom&&A[].s.y<=ytop)flag=true;
        if(A[].e.x<=xright&&A[].e.x>=xleft&&A[].e.y>=ybottom&&A[].e.y<=ytop)flag=true;
        if(i>&&flag==)  printf("F\n");
        else   printf("T\n");
    }
    return ;
}