poj 1410 Intersection

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

1
4 9 11 2 1 5 7 1

Sample Output

F

题意：给你一条线，和一个矩阵的左上和右下的点，要你求是否有交点；

吐糟：这题有2个坑点

1. 左上和右下的点要自己找出来；

2. 线段在矩形里也算有交点；

思路：我是先判断是否在矩形里面，然后在判断是否相交；

#include<iostream>
#include<cstdio>
using namespace std;
struct point
{
double x,y;
};
struct vector
{
point start,end;
};

double multi(point p1,point p2,point p0)
{
return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
int Across(vector v1,vector v2)
{
if(max(v1.start.x,v1.end.x)>=min(v2.start.x,v2.end.x)&&max(v2.start.x,v2.end.x)>=min(v1.start.x,v1.end.x)
&&max(v1.start.y,v1.end.y)>=min(v2.start.y,v2.end.y)&&max(v2.start.y,v2.end.y)>=min(v1.start.y,v1.end.y)
&&multi(v2.start,v1.end,v1.start)*multi(v1.end,v2.end,v1.start)>=0
&&multi(v1.start,v2.end,v2.start)*multi(v2.end,v1.end,v2.start)>=0)
return 1;
return 0;
}
vector line[10],l;
point p[10];
int main()
{
double x1,y1,x2,y2;
int n;
cin>>n;
for(int i=0;i<n;i++)
{
cin>>l.start.x>>l.start.y>>l.end.x>>l.end.y>>x1>>y1>>x2>>y2;
p[0].x=min(x1,x2),p[0].y=max(y1,y2);p[2].x=max(x1,x2),p[2].y=min(y1,y2);
p[1].x=p[0].x,p[1].y=p[2].y,p[3].x=p[2].x,p[3].y=p[0].y;
for(int i=0;i<4;i++)
{
line[i].start.x=p[i].x;
line[i].start.y=p[i].y;
line[i].end.x=p[(i+1)%4].x;
line[i].end.y=p[(i+1)%4].y;

}
int flag=0;
if((l.start.x==l.end.x)&&flag==0)
{
if((p[2].y<=l.start.y&&l.start.y<=p[0].y)||(p[2].y<=l.end.y&&l.end.y<=p[0].y))
{
flag=1;
printf("T\n");
}
}
if((l.start.y==l.end.y)&&flag==0)
{
if((p[0].x>=l.start.x&&l.start.x>=p[2].x)||(p[2].x>=l.end.x&&l.end.x>=p[0].x))
{
flag=1;
printf("T\n");
}
}
if(flag==0)
for(int i=0;i<4;i++)
{
if(Across(line[i],l))
{
flag=1;
printf("T\n");
break;
}
}

if((l.start.x>=p[0].x&&l.start.y<=p[0].y&&l.start.x<=p[2].x&&l.start.y>=p[2].y&&flag==0)||
(l.end.x>=p[0].x&&l.end.y<=p[0].y&&l.end.x<=p[2].x&&l.end.y>=p[2].y&&flag==0)){
flag=1;
printf("T\n");
}

if(flag==0)
printf("F\n");

}
return 0;
}
/*
9 2 9 1 4 3 9 6
9 1 9 2 4 3 9 6
*/