POJ 1410 Intersection --几何,线段相交

题意: 给一条线段,和一个矩形,问线段是否与矩形相交或在矩形内。

解法: 判断是否在矩形内,如果不在,判断与四条边是否相交即可。这题让我发现自己的线段相交函数有错误的地方,原来我写的线段相交函数就是单纯做了两次跨立实验,在下图这种情况是错误的:

《POJ 1410 Intersection --几何,线段相交》

这样的话线段与右边界的两次跨立实验(叉积<=0)都会通过,但是并不相交。

所以要加快速排斥。

还有就是这题题目说给出的不一定是左上角,右下角依次的顺序。所以干脆重新自己定义左上角,右下角。

代码:

《POJ 1410 Intersection --几何,线段相交》
《POJ 1410 Intersection --几何,线段相交》

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#define eps 1e-8
using namespace std;
#define N 100017

struct Point{
    double x,y;
    Point(double x=0, double y=0):x(x),y(y) {}
    void input() { scanf("%lf%lf",&x,&y); }
};
typedef Point Vector;
struct Circle{
    Point c;
    double r;
    Circle(){}
    Circle(Point c,double r):c(c),r(r) {}
    Point point(double a) { return Point(c.x + cos(a)*r, c.y + sin(a)*r); }
    void input() { scanf("%lf%lf%lf",&c.x,&c.y,&r); }
};
struct Line{
    Point p;
    Vector v;
    double ang;
    Line(){}
    Line(Point p, Vector v):p(p),v(v) { ang = atan2(v.y,v.x); }
    Point point(double t) { return Point(p.x + t*v.x, p.y + t*v.y); }
    bool operator < (const Line &L)const { return ang < L.ang; }
};
int dcmp(double x) {
    if(x < -eps) return -1;
    if(x > eps) return 1;
    return 0;
}
template <class T> T sqr(T x) { return x * x;}
Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }
Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }
Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }
bool operator < (const Point& a, const Point& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
bool operator >= (const Point& a, const Point& b) { return a.x >= b.x && a.y >= b.y; }
bool operator <= (const Point& a, const Point& b) { return a.x <= b.x && a.y <= b.y; }
bool operator == (const Point& a, const Point& b) { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; }
double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
double Length(Vector A) { return sqrt(Dot(A, A)); }
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
Vector VectorUnit(Vector x){ return x / Length(x);}
Vector Normal(Vector x) { return Point(-x.y, x.x) / Length(x);}
double angle(Vector v) { return atan2(v.y, v.x); }

bool SegmentIntersection(Point A,Point B,Point C,Point D) {
    return max(A.x,B.x) >= min(C.x,D.x) &&
           max(C.x,D.x) >= min(A.x,B.x) &&
           max(A.y,B.y) >= min(C.y,D.y) &&
           max(C.y,D.y) >= min(A.y,B.y) &&
           dcmp(Cross(C-A,B-A)*Cross(D-A,B-A)) <= 0 &&
           dcmp(Cross(A-C,D-C)*Cross(B-C,D-C)) <= 0;
}

//data segment
struct node{
    Point P[2];
}line[206];
//data ends

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        double x1,y1,x2,y2;
        double xleft,ytop,xright,ybottom;
        Point A,B;
        A.input(), B.input();
        scanf("%lf%lf%lf%lf",&xleft,&ytop,&xright,&ybottom);
        Point P1,P2,P3,P4;
        double XL = min(xleft,xright);
        double XR = max(xleft,xright);
        double YB = min(ybottom,ytop);
        double YT = max(ybottom,ytop);
        xleft = XL, xright = XR, ybottom = YB, ytop = YT;
        P1 = Point(xleft,ytop);
        P2 = Point(xleft,ybottom);
        P3 = Point(xright,ytop);
        P4 = Point(xright,ybottom);
        int flag = 0;
        if(SegmentIntersection(A,B,P1,P3) || SegmentIntersection(A,B,P1,P2) || SegmentIntersection(A,B,P2,P4) || SegmentIntersection(A,B,P3,P4))
            flag = 1;
        if(dcmp(A.x-xleft) >= 0 && dcmp(A.x-xright) <= 0 && dcmp(A.y-ytop) <= 0 && dcmp(A.y-ybottom) >= 0 && dcmp(B.x-xleft) >= 0 && dcmp(B.x-xright) <= 0 && dcmp(B.y-ytop) <= 0 && dcmp(B.y-ybottom) >= 0)
            flag = 1;
        if(flag) puts("T");
        else     puts("F");
    }
    return 0;
}

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    原文作者:hq572241670
    原文地址: https://blog.csdn.net/hq572241670/article/details/41550845
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