题目:传送门
题意:有 n 个测试样例,每个样例,输入四个点,前两个点代表一条线段,后两个点代表正方形的两个对角端点。
#include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <queue> #include <map> #include <vector> #include <set> #include <string> #include <math.h> #define LL long long #define mem(i, j) memset(i, j, sizeof(i)) #define rep(i, j, k) for(int i = j; i <= k; i++) #define dep(i, j, k) for(int i = k; i >= j; i--) #define pb push_back #define make make_pair #define INF 1e20 #define inf LLONG_MAX #define PI acos(-1) using namespace std; const int N = 1e2 + 5; const double eps = 1e-10; struct Point { double x, y; Point(double x = 0, double y = 0) : x(x), y(y) { } /// 构造函数 }; /// 向量加减乘除 inline Point operator + (const Point& A, const Point& B) { return Point(A.x + B.x, A.y + B.y); } inline Point operator - (const Point& A, const Point& B) { return Point(A.x - B.x, A.y - B.y); } inline Point operator * (const Point& A, const double& p) { return Point(A.x * p, A.y * p); } inline Point operator / (const Point& A, const double& p) { return Point(A.x / p, A.y / p); } inline int dcmp(const double& x) { if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; } inline double Cross(const Point& A, const Point& B) { return A.x * B.y - A.y * B.x; } /// 叉积 inline double Dot(const Point& A, const Point& B) { return A.x * B.x + A.y * B.y; } /// 点积 inline double Length(const Point& A) { return sqrt(Dot(A, A)); } /// 向量长度 inline double Angle(const Point& A, const Point& B) { return acos(Dot(A, B) / Length(A) / Length(B)); } /// 向量A,B夹角 inline Point GetLineIntersection(const Point P, const Point v, const Point Q, const Point w) {///求直线p + v*t 和 Q + w*t 的交点,需确保有交点,v和w是方向向量 Point u = P - Q; double t = Cross(w, u) / Cross(v, w); return P + v * t; } inline bool Onsegment(Point p, Point a1, Point a2) { /// 判断点p是否在线段p1p2上 return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) <= 0; } inline bool SegmentProperInsection(Point a1, Point a2, Point b1, Point b2) { /// 判断线段是否相交 if(dcmp(Cross(a1 - a2, b1 - b2)) == 0) /// 两线段平行 return Onsegment(b1, a1, a2) || Onsegment(b2, a1, a2) || Onsegment(a1, b1, b2) || Onsegment(a2, b1, b2); Point tmp = GetLineIntersection(a1, a2 - a1, b1, b2 - b1); return Onsegment(tmp, a1, a2) && Onsegment(tmp, b1, b2); } inline int isPointInpolygon(Point tmp, Point P[], int n) { /// 判断点是否在多边形里 int wn = 0; rep(i, 0, n - 1) { if(Onsegment(tmp, P[i], P[(i + 1) % n])) return -1; /// 边界 int k = dcmp(Cross(P[(i + 1) % n] - P[i], tmp - P[i])); int d1 = dcmp(P[i].y - tmp.y); int d2 = dcmp(P[(i +1) % n].y - tmp.y); if(k > 0 && d1 <= 0 && d2 > 0) wn++; if(k < 0 && d2 <= 0 && d1 > 0) wn--; } if(wn) return 1; /// 外部 return 0; /// 内部 } Point P[N]; void solve() { Point st, ed; double x1, x2, y1, y2; scanf("%lf %lf %lf %lf", &st.x, &st.y, &ed.x, &ed.y); scanf("%lf %lf %lf %lf", &x1, &y1, &x2, &y2); if(x1 > x2) swap(x1, x2); if(y1 > y2) swap(y1, y2); P[0] = Point(x1, y1); P[1] = Point(x1, y2); P[2] = Point(x2, y2); P[3] = Point(x2, y1); rep(i, 0, 2) { /// 判断线段是否和多边形的边相交 if(SegmentProperInsection(st, ed, P[i], P[i + 1]) == 1) { puts("T"); return ; } } if(isPointInpolygon(st, P, 4) || isPointInpolygon(ed, P, 4)) { /// 判断线段是否有一个端点在多边形里或者边界上 puts("T"); return ; } puts("F"); } int main() { int _; scanf("%d", &_); while(_--) solve(); return 0; }
原文地址:https://www.cnblogs.com/Willems/p/12381126.html
时间: 2024-08-02 21:53:34