Intersection （计算几何）

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.

果然需要积累模板...一下子就用到了

简单说就是先判断线段是不是在矩形内，再分别判断和四条边是否相交...

判断相交貌似一直是比较棘手的问题啊，不过这种线线相交还好...

#include<cstdio>#include<cstring>#include<cmath>#define eps 1e-8#define zero(x) (((x)>0?(x):-(x))<eps)struct point{    double x, y;};double xmult(point p1, point p2, point p0){    return (p1.x - p0.x)*(p2.y - p0.y) - (p2.x - p0.x)*(p1.y - p0.y);}double dmult(point p1, point p2, point p0){    return (p1.x - p0.x)*(p2.x - p0.x) + (p1.y - p0.y)*(p2.y - p0.y);}int dots_inline(point p1, point p2, point p3){    return zero(xmult(p1, p2, p3));}int dot_online_in(point p, point l1, point l2){    return zero(xmult(p, l1, l2)) && (l1.x - p.x)*(l2.x - p.x)<eps && (l1.y - p.y)*(l2.y - p.y)<eps;}int same_side(point p1, point p2, point l1, point l2){    return xmult(l1, p1, l2)*xmult(l1, p2, l2)>eps;}int intersect_in(point u1, point u2, point v1, point v2){    if (!dots_inline(u1, u2, v1) || !dots_inline(u1, u2, v2))        return !same_side(u1, u2, v1, v2) && !same_side(v1, v2, u1, u2);    return dot_online_in(u1, v1, v2) || dot_online_in(u2, v1, v2) || dot_online_in(v1, u1, u2) || dot_online_in(v2, u1, u2);}int main(){    point x, y;    point d1, d2, d3, d4;    int cas, i, j, k;    scanf("%d", &cas);    while (cas--)    {        scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &x.x, &x.y, &y.x, &y.y, &d4.x, &d4.y, &d2.x, &d2.y);        d1.x = d4.x, d1.y = d2.y;        d3.x = d2.x, d3.y = d4.y;        if ((x.x >= d2.x&&x.x <= d4.x) || (x.x >= d4.x&&x.x <= d2.x))            if ((x.y >= d2.y&&x.y <= d4.y) || (x.y >= d4.y&&x.y <= d2.y))            {                printf("T\n");                continue;            }        if ((y.x >= d2.x&&y.x <= d4.x) || (y.x >= d4.x&&y.x <= d2.x))            if ((y.y >= d2.y&&y.y <= d4.y) || (y.y >= d4.y&&y.y <= d2.y))            {                printf("T\n");                continue;            }        if (intersect_in(x, y, d1, d2) || intersect_in(x, y, d3, d2) || intersect_in(x, y, d3, d4) || intersect_in(x, y, d1, d4))            printf("T\n");        else printf("F\n");    }    return 0;}