POJ-1410-Intersection（线段相交）

题目链接：http://poj.org/problem?id=1410

题目大意：给出一个线段和一个矩形，判断他们是否相交

思路：因为只有几种情况，全部列出来就好了，注意，线段全部在矩形中不算相交。

ACCode：

//#pragma comment(linker, "/STACK:1024000000,1024000000")
  
#include<stdio.h>
#include<string.h> 
//#include<math.h> 
   
#include<map>  
#include<set>
#include<cmath>
#include<deque> 
#include<queue> 
#include<stack> 
#include<bitset>
#include<string> 
#include<fstream>
#include<iostream> 
#include<algorithm> 
using namespace std; 
  
#define ll long long 
#define Pair pair<int,int>
//#define max(a,b) (a)>(b)?(a):(b)
//#define min(a,b) (a)<(b)?(a):(b)
#define clean(a,b) memset(a,b,sizeof(a))// ??
//std::ios::sync_with_stdio(false);
//  register
const int MAXN=1e2+10;
const int INF32=0x3f3f3f3f;
const ll INF64=0x3f3f3f3f3f3f3f3f;
const ll mod=1e9+7;
const double PI=acos(-1.0);
const double EPS=1.0e-8;

struct Point{
	double x,y,t,d;
	Point(double _x=0,double _y=0,double _t=0,double _d=0){
		x=_x;y=_y;t=_t;d=_d;
	}
	friend Point operator + (const Point &a,const Point &b){
		return Point(a.x+b.x,a.y+b.y);
	}
	friend Point operator - (const Point &a,const Point &b){
		return Point(a.x-b.x,a.y-b.y);
	}
	friend double operator ^ (Point a,Point b){//??????? 
		return a.x*b.y-a.y*b.x;
	}
	friend int operator == (const Point &a,const Point &b){
		if(fabs(a.x-b.x)<EPS&&fabs(a.y-b.y)<EPS) return 1;
		return 0;
	}
};
struct V{
	Point start,end;double ang;
	V(Point _start=Point(0,0),Point _end=Point(0,0),double _ang=0.0){
		start=_start;end=_end;ang=_ang;
	}
	friend V operator + (const V &a,const V &b){
		return V(a.start+b.start,a.end+b.end);
	}
	friend V operator - (const V &a,const V &b){
		return V(a.start-b.start,a.end-b.end);
	}
};
V Line[MAXN];
double X1,X2,Y1,Y2;
int n;

int LineInter(V l1,V l2){
	if(max(l1.start.x,l1.end.x)>=min(l2.start.x,l2.end.x)&&
	max(l2.start.x,l2.end.x)>=min(l1.start.x,l1.end.x)&&
	max(l1.start.y,l1.end.y)>=min(l2.start.y,l2.end.y)&&
	max(l2.start.y,l2.end.y)>=min(l1.start.y,l1.end.y)){
		if(((l2.end-l2.start)^(l1.start-l2.start))*((l2.end-l2.start)^(l1.end-l2.start))<=0&&
		((l1.end-l1.start)^(l2.start-l1.start))*((l1.end-l1.start)^(l2.end-l1.start))<=0)
			//printf(" 相交： l1:(%lf,%lf),(%lf,%lf) l2:(%lf,%lf),(%lf,%lf)\n",l1.start.x,l1.start.y,l1.end.x,l1.end.y,l2.start.x,l2.start.y,l2.end.x,l2.end.y);
			return 1;
	}//printf("不相交： l1:(%lf,%lf),(%lf,%lf) l2:(%lf,%lf),(%lf,%lf)\n",l1.start.x,l1.start.y,l1.end.x,l1.end.y,l2.start.x,l2.start.y,l2.end.x,l2.end.y);
	return 0;
}
int InRec(Point p){
	if(p.x<X2+EPS&&p.x>X1-EPS){
		if(p.y<Y1+EPS&&p.y>Y2-EPS){
			return 1;
		}
	}return 0;
}
int Judge(){
	return InRec(Line[0].start)||InRec(Line[0].end);
}
int main(){
	scanf("%d",&n);
	for(int j=1;j<=n;++j){
		double x1,x2,y1,y2;
		scanf("%lf%lf%lf%lf",&X1,&Y1,&X2,&Y2);
		x1=X1,x2=X2,y1=Y1,y2=Y2;
		Line[0]=V(Point(x1,y1),Point(x2,y2));
		scanf("%lf%lf%lf%lf",&X1,&Y1,&X2,&Y2);
		x1=min(X1,X2),x2=max(X1,X2),y1=max(Y1,Y2),y2=min(Y1,Y2);
		X1=x1,X2=x2,Y1=y1,Y2=y2;
		Line[1]=V(Point(x1,y1),Point(x2,y1));
		Line[2]=V(Point(x1,y2),Point(x2,y2));
		Line[3]=V(Point(x1,y1),Point(x1,y2));
		Line[4]=V(Point(x2,y1),Point(x2,y2));
		int flag=0;
		if(Judge()) flag=1;
		for(int i=1;i<=4;++i){
			if(LineInter(Line[0],Line[i])){
				flag=1;break;
			}
		}
		if(flag) puts("T");
		else puts("F");
	}
}