POJ - 1410 - Intersection（计算几何）

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

题目链接
计算几何
这是一个计算几何的模板题目，可以看一下计算几何的板子，里面也讲得特别详细，主要是通过快速排斥和跨历实验来判断两条线段是否相交。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define eps 1e-7
using namespace std;
int n;

//结构体记录点
struct Node
{
    double x, y;
    Node(double x = 0, double y = 0) : x(x), y(y) {}
    Node operator - (const Node& obj) const
    {
        return Node(x - obj.x, y - obj.y);
    }
    //叉乘
    double operator * (const Node& obj) const
	{
	    return x * obj.y - y * obj.x;
	}
}line_start, line_end, vertex[6];

int accuracy(double x)  //精度问题处理
{
	if (fabs(x) < eps) return 0;
	return x < 0 ? -1 : 1;
}

//判断是否相交
int inter(Node point1_line1, Node point2_line1, Node point1_line2, Node point2_line2)
{
    if(min(point1_line1.x, point2_line1.x) > max(point1_line2.x, point2_line2.x) || min(point1_line1.y, point2_line1.y) > max(point1_line2.y, point2_line2.y))
        return 0;
    if(min(point1_line2.x, point2_line2.x) > max(point1_line1.x, point2_line1.x) || min(point1_line2.y, point2_line2.y) > max(point1_line1.y, point2_line1.y))
        return 0;
    double val1 = (point2_line1 - point1_line1) * (point1_line2 - point1_line1);
    double val2 = (point2_line1 - point1_line1) * (point2_line2 - point1_line1);
    double val3 = (point2_line2 - point1_line2) * (point1_line1 - point1_line2);
    double val4 = (point2_line2 - point1_line2) * (point2_line1 - point1_line2);
    return accuracy(val1) * accuracy(val2) <= 0 && accuracy(val3) * accuracy(val4) <= 0;
}

判断点是否在矩形内部
int inside(Node now)
{
	double x = now.x; double y = now.y;
	return x >= vertex[1].x && x <= vertex[3].x && y <= vertex[1].y && y >= vertex[3].y;
}

int main()
{
    scanf("%d", &n);
    while(n--)
    {
        scanf("%lf%lf", &line_start.x, &line_start.y);
        scanf("%lf%lf", &line_end.x, &line_end.y);
        scanf("%lf%lf", &vertex[1].x, &vertex[1].y);
        scanf("%lf%lf", &vertex[3].x, &vertex[3].y);
        if(vertex[1].x > vertex[3].x)   swap(vertex[1].x, vertex[3].x);
        if(vertex[1].y < vertex[3].y)   swap(vertex[1].y, vertex[3].y);
        vertex[2].x = vertex[3].x;  vertex[2].y = vertex[1].y;
        vertex[4].x = vertex[1].x;  vertex[4].y = vertex[3].y;
        int ans = 0;
        ans += inter(vertex[1], vertex[2], line_start, line_end);
        ans += inter(vertex[2], vertex[3], line_start, line_end);
        ans += inter(vertex[3], vertex[4], line_start, line_end);
        ans += inter(vertex[4], vertex[1], line_start, line_end);
        if(inside(line_start) || inside(line_end))  ans++;
        if(ans) printf("T\n");
        else    printf("F\n");
    }
    return 0;
}