湖南大学ACM程序设计新生杯大赛（同步赛）C-Do you like Banana ? 【几何】

时间限制：C/C++ 1秒，其他语言2秒

空间限制：C/C++ 32768K，其他语言65536K

64bit IO Format: %lld

题目描述

Two endpoints of two line segments on a plane are given to determine whether the two segments are intersected (there is a common point or there is a partial coincidence that intersects). If intersected, output “Yes”, otherwise output “No”.

输入描述:

The first line is a number of T, indicating the number of tests inputed (1 <= T <= 1000)

For next T line,each line contains 8 numbers , x1,y1,x2,y2,x3,y3,x4, y4. (8-10 ^ < = xi, yi < = 10 ^ 8)

(the two endpoints of line 1 are x1, y1, |, x2, y2, and two of the endpoints of line 2 are x3, y3, |, x4, y4).

输出描述:

For each test case, output”Yes” if the two segments intersected, else output”No”.

示例1

输入

2

1 2 2 1 0 0 2 2

-1 1 1 1 0 0 1 -1

输出

Yes

No

问题：求两个线段能否相交。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
///----------alg 1------------
struct Point
{
double x, y;
};

bool between(double a, double X0, double X1)
{
double temp1 = a - X0;
double temp2 = a - X1;
if ((temp1<1e-8 && temp2>-1e-8) || (temp2<1e-6 && temp1>-1e-8))
{
return true;
}
else
{
return false;
}
}

// 判断两条直线段是否有交点，有则计算交点的坐标
// p1,p2是直线一的端点坐标
// p3,p4是直线二的端点坐标
bool detectIntersect(Point p1, Point p2, Point p3, Point p4)
{
double line_x, line_y; //交点
if ((fabs(p1.x - p2.x)<1e-6) && (fabs(p3.x - p4.x)<1e-6))
{
return false;
}
else if ((fabs(p1.x - p2.x)<1e-6)) //如果直线段p1p2垂直与y轴
{
if (between(p1.x, p3.x, p4.x))
{
double k = (p4.y - p3.y) / (p4.x - p3.x);
line_x = p1.x;
line_y = k*(line_x - p3.x) + p3.y;

if (between(line_y, p1.y, p2.y))
{
return true;
}
else
{
return false;
}
}
else
{
return false;
}
}
else if ((fabs(p3.x - p4.x)<1e-6)) //如果直线段p3p4垂直与y轴
{
if (between(p3.x, p1.x, p2.x))
{
double k = (p2.y - p1.y) / (p2.x - p1.x);
line_x = p3.x;
line_y = k*(line_x - p2.x) + p2.y;

if (between(line_y, p3.y, p4.y))
{
return true;
}
else
{
return false;
}
}
else
{
return false;
}
}
else
{
double k1 = (p2.y - p1.y) / (p2.x - p1.x);
double k2 = (p4.y - p3.y) / (p4.x - p3.x);

if (fabs(k1 - k2)<1e-6)
{
return false;
}
else
{
line_x = ((p3.y - p1.y) - (k2*p3.x - k1*p1.x)) / (k1 - k2);
line_y = k1*(line_x - p1.x) + p1.y;
}

if (between(line_x, p1.x, p2.x) && between(line_x, p3.x, p4.x))
{
return true;
}
else
{
return false;
}
}
}
///------------alg 1------------

int main()
{
int T;
scanf("%d", &T);
while (T--) {
double x1, x2, x3, x4;
double y1, y2, y3, y4;
scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &x1, &y1, &x2, &y2, &x3, &y3, &x4, &y4);
Point p1, p2, p3, p4;
p1.x = x1, p1.y = y1;
p2.x = x2, p2.y = y2;
p3.x = x3, p3.y = y3;
p4.x = x4, p4.y = y4;
if (detectIntersect(p1, p2, p3, p4)) {
printf("Yes\n");
}
else
{
printf("No\n");
}
}
return 0;
}