POJ1410Intersection【判断线段与矩形相交+点在矩形内的简单判定】

Language:Default Intersection Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12880   Accepted: 3362 Description You are to write a program that has to decide whether a given line segment intersects a given rectangle.  An example:  line: start point: (4,9)  end point: (11,2)  rectangle: left-top: (1,5)  right-bottom: (7,1)    Figure 1: Line segment does not intersect rectangle  The line is said to intersect the rectangle if the line and the rectangle have at least one point in common. The rectangle consists of four straight lines and the area in between. Although all input values are integer numbers, valid intersection points do not have to lay on the integer grid.  Input The input consists of n test cases. The first line of the input file contains the number n. Each following line contains one test case of the format:  xstart ystart xend yend xleft ytop xright ybottom  where (xstart, ystart) is the start and (xend, yend) the end point of the line and (xleft, ytop) the top left and (xright, ybottom) the bottom right corner of the rectangle. The eight numbers are separated by a blank. The terms top left and bottom right do not imply any ordering of coordinates. Output For each test case in the input file, the output file should contain a line consisting either of the letter "T" if the line segment intersects the rectangle or the letter "F" if the line segment does not intersect the rectangle. Sample Input 14 9 11 2 1 5 7 1 Sample Output F

题意：判断一条线段是否和矩形相交；

想到了直线可能完全在矩形内部但却没有完全理解题意The terms top left and bottom right do not imply any ordering of coordinates；注意它们的位置关系不确定

#include
#include
#include
#define eps 1e-8
using namespace std;
struct point{
	double x,y;
};
struct line{
	point a,b;
}A[100010];
double MAX(double a,double b){
	return a>b?a:b;
}
double MIN(double a,double b){
	return aMAX(A[b].a.x,A[b].b.x)||MIN(A[a].a.y,A[a].b.y)>MAX(A[b].a.y,A[b].b.y)||MIN(A[b].a.x,A[b].b.x)>MAX(A[a].a.x,A[a].b.x)||MIN(A[b].a.y,A[b].b.y)>MAX(A[a].a.y,A[a].b.y))      
    return false;      
    double h,i,j,k;      
    h=(A[a].b.x-A[a].a.x)*(A[b].a.y-A[a].a.y)-(A[a].b.y-A[a].a.y)*(A[b].a.x-A[a].a.x);      
    i=(A[a].b.x-A[a].a.x)*(A[b].b.y-A[a].a.y)-(A[a].b.y-A[a].a.y)*(A[b].b.x-A[a].a.x);      
    j=(A[b].b.x-A[b].a.x)*(A[a].a.y-A[b].a.y)-(A[b].b.y-A[b].a.y)*(A[a].a.x-A[b].a.x);      
    k=(A[b].b.x-A[b].a.x)*(A[a].b.y-A[b].a.y)-(A[b].b.y-A[b].a.y)*(A[a].b.x-A[b].a.x);      
    return h*i<=eps&&j*k<=eps;      
}      
int main()
{
	int t,i,j,k;
	double xleft,ytop,xright,ybottom;
	double x1,y1,x2,y2;
	scanf("%d",&t);
	while(t--){
		scanf("%lf%lf%lf%lf",&A[0].a.x,&A[0].a.y,&A[0].b.x,&A[0].b.y);
		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].a.x=xleft;A[1].a.y=ybottom;A[1].b.x=xleft;A[1].b.y=ytop;
		A[2].a.x=xleft;A[2].a.y=ytop;A[2].b.x=xright;A[2].b.y=ytop;
		A[3].a.x=xright;A[3].a.y=ytop;A[3].b.x=xright;A[3].b.y=ybottom;
		A[4].a.x=xright;A[4].a.y=ybottom;A[4].b.x=xleft;A[4].b.y=ybottom;
		for(i=1;i<=4;++i){
			if(judge(0,i))break;
		}
		bool flag=false;
		if(A[0].a.x<=xright&&A[0].a.x>=xleft&&A[0].a.y>=ybottom&&A[0].a.y<=ytop)flag=true;
		if(A[0].b.x<=xright&&A[0].b.x>=xleft&&A[0].b.y>=ybottom&&A[0].b.y<=ytop)flag=true; 
		if(i>4&&flag==0)
			printf("F\n");
		else 
			printf("T\n");
	}
	return 0;
}