标签:false lse queue intersect 叉积 思维 point using ||
题目: 传送门
题意: 给你n条线段的两个端点,问所有线段投影到一条直线上,这些投影至少相交于一点,就输出Yes!,否则就是 No!
题解: 戳
#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 INT_MAX #define inf LLONG_MAX #define PI acos(-1) using namespace std; const int N = 310; struct Point { double x, y; Point(double x = 0, double y = 0) : x(x), y(y) { } /// 构造函数 }; typedef Point Vector; /// 向量+向量=向量, 点+向量=向量 Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); } ///点-点=向量 Vector operator - (Point A, Point B) { return Vector(A.x - B.x, A.y - B.y); } ///向量*数=向量 Vector operator * (Vector A, double p) { return Vector(A.x * p, A.y * p); } ///向量/数=向量 Vector operator / (Vector A, double p) { return Vector(A.x / p, A.y / p); } const double eps = 1e-10; int dcmp(double x) { if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; } bool operator < (const Point& a, const Point& b) { return a.x == b.x ? a.y < b.y : a.x < b.x; } 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; } /// 点积 double Length(Vector A) { return sqrt(Dot(A, A)); } /// 计算向量长度 double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); } /// 向量A、B夹角 double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; } /// 叉积 Vector Rotate(Vector A, double rad) { /// 向量旋转, rad 是弧度, 逆时针旋转 rad return Vector(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad)); } Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) { /// 求两直线交点,请确保 P + tv 和 Q + tw 有唯一交点。 当且仅当Cross(v, w)非0 Vector u = P - Q; double t = Cross(w, u) / Cross(v, w); return P + v * t; } bool SegmentProperInsection(Point a1, Point a2, Point b1, Point b2) { /// 判断线段是否相交 double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1); double c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1); return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0; } bool OnSegment(Point p, Point a1, Point a2) { /// 点p是否在线段a1,a2上(不包含端点) return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0; } bool LSProperInsection(Point a1, Point a2, Point b1, Point b2) { /// 判断直线b1b2是否和线段a1a2相交 return dcmp(Cross(b1 - a1, b2 - a1)) * dcmp(Cross(b1 - a2, b2 - a2)) <= 0; } Point P[N], Q[N]; int n; bool judge(Point a, Point b) { if(dcmp(Length(b - a)) == 0) return false; rep(i, 1, n) if(LSProperInsection(P[i], Q[i], a, b) == false) return false; return true; } void solve() { scanf("%d", &n); rep(i, 1, n) scanf("%lf %lf %lf %lf", &P[i].x, &P[i].y, &Q[i].x, &Q[i].y); rep(i, 1, n) rep(j, 1, n) if(judge(P[i], P[j]) || judge(P[i], Q[j]) || judge(Q[i], P[j]) || judge(Q[i], Q[j])) { puts("Yes!"); return ; } puts("No!"); return ; } int main() { int _; scanf("%d", &_); while(_--) solve(); return 0; }
POJ 3304 Segments (线段和直线相交 + 思维)
标签:false lse queue intersect 叉积 思维 point using ||
原文地址:https://www.cnblogs.com/Willems/p/12337434.html