[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[35];
int sgn(double x){
    if(fabs(x)<eps)return 0;
    else if(x<0) return -1;
    else return 1;
}
//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))<=0&&
        sgn((l1.s-l2.s)^(l2.e-l2.s))*sgn((l1.e-l2.s)^(l2.e-l2.s))<=0;
} 

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[0].s.x,&A[0].s.y,&A[0].e.x,&A[0].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[1].s.x=xleft;A[1].s.y=ybottom;A[1].e.x=xleft;A[1].e.y=ytop;  
        A[2].s.x=xleft;A[2].s.y=ytop;A[2].e.x=xright;A[2].e.y=ytop;  
        A[3].s.x=xright;A[3].s.y=ytop;A[3].e.x=xright;A[3].e.y=ybottom;  
        A[4].s.x=xright;A[4].s.y=ybottom;A[4].e.x=xleft;A[4].e.y=ybottom;//矩形的四条线段 
        for(i=1;i<=4;++i) if(inter(A[0],A[i]))break;
        bool flag=false;//矩形是实心的。 
        if(A[0].s.x<=xright&&A[0].s.x>=xleft&&A[0].s.y>=ybottom&&A[0].s.y<=ytop)flag=true;  
        if(A[0].e.x<=xright&&A[0].e.x>=xleft&&A[0].e.y>=ybottom&&A[0].e.y<=ytop)flag=true;   
        if(i>4&&flag==0)  printf("F\n");
        else   printf("T\n");  
    }  
    return 0;  
}