题目链接:POJ 1410 Intersection
判断一条线段是否与一个矩形有交点,如果线段在矩形内部,也算有交点。
这题挺多细节需要考虑,要不是discuss上给的68组数据,我还得用更多时间去改bug。
有两个点在矩形内部比较容易处理,面积法搞定,难点在于相交的处理。刘汝佳的模版是线段规范相交的,而这道题是非规范的,就是说可能一条线段的端点恰好与另一条线段相交。但是仅仅加上OnSegment也是不行的,因为这个函数不能处理两个线段恰好端点相交这种情况。
#include <iostream> #include <cstdio> #include <cmath> using namespace std; const double eps = 1e-10; struct Point { int x, y; Point(int x=0, int y=0):x(x),y(y) { } }; struct Line { Point a, b; }; typedef Point Vector; Vector operator + (const Vector& A, const Vector& B) { return Vector(A.x+B.x, A.y+B.y); } Vector operator - (const Point& A, const Point& B) { return Vector(A.x-B.x, A.y-B.y); } Vector operator * (const 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) == 0 && (a.y-b.y) == 0; } double Cross(const Vector& A, const Vector& B) { return A.x*B.y - A.y*B.x; } bool SegmentProperIntersection(const Point& a1, const Point& a2, const Point& b1, const Point& b2) { int c1 = Cross(a2-a1,b1-a1), c2 = Cross(a2-a1,b2-a1), c3 = Cross(b2-b1,a1-b1), c4=Cross(b2-b1,a2-b1); return c1*c2<0 && c3*c4<0; } int Dot(const Vector& A, const Vector& B) { return A.x*B.x + A.y*B.y; } bool OnSegment(const Point& p, const Point& a1, const Point& a2) { return Cross(a1-p, a2-p) == 0 && Dot(a1-p, a2-p) < 0; } Point p[4]; int get_area(Point a, Point b, Point c) { return abs(a.x * b.y + c.x * a.y + b.x * c.y - c.x * b.y - a.x * c.y - b.x * a.y); } bool is_in(int sum, Point k1, Point k2) { double area1, area2, area3, area4; area1 = get_area(p[0], p[1], k1); area2 = get_area(p[1], p[2], k1); area3 = get_area(p[2], p[3], k1); area4 = get_area(p[3], p[0], k1); if(2 * sum == area1 + area2 + area3 + area4) return true; area1 = get_area(p[0], p[1], k2); area2 = get_area(p[1], p[2], k2); area3 = get_area(p[2], p[3], k2); area4 = get_area(p[3], p[0], k2); if(2 * sum == area1 + area2 + area3 + area4) return true; return false; } bool is_intersect(Point k1, Point k2) { if(SegmentProperIntersection(k1, k2, p[0], p[1])) return true; if(SegmentProperIntersection(k1, k2, p[1], p[2])) return true; if(SegmentProperIntersection(k1, k2, p[2], p[3])) return true; if(SegmentProperIntersection(k1, k2, p[3], p[0])) return true; if(OnSegment(k1, p[0], p[1]) || OnSegment(k1, p[1], p[2]) || OnSegment(k1, p[2], p[3]) || OnSegment(k1, p[3], p[0])) return true; if(OnSegment(k2, p[0], p[1]) || OnSegment(k2, p[1], p[2]) || OnSegment(k2, p[2], p[3]) || OnSegment(k2, p[3], p[0])) return true; if(OnSegment(p[0], k1, k2) || OnSegment(p[1], k1, k2) || OnSegment(p[2], k1, k2) || OnSegment(p[3], k1, k2)) return true; if(k1 == p[0] || k1 == p[1] || k1 == p[2] || k1 == p[3]) return true; if(k2 == p[0] || k2 == p[1] || k2 == p[2] || k2 == p[3]) return true; return false; } int main() { //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); int T; scanf("%d", &T); Line line; while(T--) { scanf("%d%d%d%d", &line.a.x, &line.a.y, &line.b.x, &line.b.y); int lx, rx, by, ty, temp; scanf("%d%d%d%d", &lx, &ty, &rx, &by); if(lx > rx) temp = lx, lx = rx, rx = temp; if(by > ty) temp = by, by = ty, ty = temp; p[0] = Point(lx, ty), p[1] = Point(rx, ty), p[2] = Point(rx, by), p[3] = Point(lx, by); int area = (ty - by) * (rx - lx); if(is_in(area, line.a, line.b) || is_intersect(line.a, line.b)) printf("T\n"); else printf("F\n"); } return 0; }