因为a,b是同时出发,同时达到,但是他们的速度不一定一样,所以我们可以设他们的速度为La,Lb(La,Lb为a,b狗的总路程)
那么,如何a,b都是沿着直线运动的时候如何求他们之间的最短最长距离呢?
因为运动都是相对的,所以我们可以把a看成不懂的,而b相对于a的移动方向就是Vb - Va,因此就可以看成a和线段 b + (Vb -Va)之间的关系了
至于方向Va,Vb向量怎么求,我们可以利用单位向量 X 移动时间 X 移动速度 得到。
经典的一道题,顺便还修复了一个模板的BUG:
14128183 | 11796 | Dog Distance | Accepted | C++ | 0.039 | 2014-09-01 09:14:46 |
#include <iostream> #include <cstdlib> #include <cstdio> #include <string> #include <cstring> #include <cmath> #include <vector> #include <queue> #include <stack> #include <algorithm> using namespace std; const double eps = 1e-10; #define MAXD 50 + 10 #define _PI acos(-1.0) struct Point { double x, y; Point(double x = 0, double y = 0) : x(x), y(y) { } bool operator < (const Point& a) const { if(a.x != x) return x < a.x; return y < a.y; } }; typedef Point Vector; Vector operator + (Vector A, Vector B) { return Vector(A.x+B.x, A.y+B.y); } Vector operator - (Point A, Point 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); } 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 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 Dist(Point A,Point B){return sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (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; } double Area2(Point A, Point B, Point C) { return Cross(B-A, C-A); } Vector Rotate(Vector A, double rad) { return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad)); } double DistToSegment(Point P,Point A,Point B){ if(A == B) return Length(P - A); Vector v1 = B - A , v2 = P - A , v3 = P - B; if(dcmp(Dot(v1,v2)) < 0) return Length(v2); if(dcmp(Dot(v1,v3)) > 0) return Length(v3); return fabs(Cross(v1,v2) / Length(v1)); } Point GetIntersection(Point P, Vector v, Point Q, Vector w) { Vector u = P-Q; double t = Cross(w, u) / Cross(v, w); return P+v*t; } bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) { 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; } double PolygonArea(Point* p, int n) { double area = 0; for(int i = 1; i < n-1; i++) area += Cross(p[i]-p[0], p[i+1]-p[0]); return area; } bool OnSegment(Point p, Point a1, Point a2) { return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0; } Point Pa[MAXD],Pb[MAXD]; int na,nb; double La,Lb; double Min ,Max; void init(){ scanf("%d%d",&na,&nb); La = 0; Lb = 0; Min = 1e20; Max = 0; for(int i = 0 ; i < na ; i++){ scanf("%lf%lf",&Pa[i].x,&Pa[i].y); if(i > 0){ La += Length(Pa[i] - Pa[i - 1]); //La += Dist(Pa[i] , Pa[i - 1]); } } for(int i = 0 ; i < nb ; i++){ scanf("%lf%lf",&Pb[i].x,&Pb[i].y); if(i > 0){ Lb += Length(Pb[i] - Pb[i - 1]); //Lb += Dist(Pb[i] , Pb[i - 1]); } } } void update(Point p,Point a,Point b){ Min = min(Min,DistToSegment(p,a,b)); Max = max(Max, Length(p - a)); Max = max(Max, Length(p - b)); } double solve(){ //假设a、b狗的速度分别为La、Lb Point pos_a = Pa[0];//a狗当前位置 Point pos_b = Pb[0];//b狗当前位置 for(int next_a = 0,next_b = 0 ; next_a < na - 1 && next_b < nb - 1; ){ double Ta = Dist(Pa[next_a + 1],pos_a); //a到下一个拐点的距离 double Tb = Dist(Pb[next_b + 1],pos_b); //b到下一个拐点的距离 double T = min(Ta / La,Tb / Lb); //最短时间 Vector Va = (Pa[next_a + 1] - pos_a) / Ta * T * La; Vector Vb = (Pb[next_b + 1] - pos_b) / Tb * T * Lb; update(pos_a,pos_b,pos_b + Vb - Va); pos_a = pos_a + Va; pos_b = pos_b + Vb; if(pos_a == Pa[next_a + 1]) next_a ++; if(pos_b == Pb[next_b + 1]) next_b ++; } //printf("%lf %lf\n",Max,Min); return (Max - Min); } int main(){ int T,Case = 1; scanf("%d",&T); while(T--){ init(); double dist = solve(); printf("Case %d: %.0f\n",Case++,dist); } return 0; }
时间: 2024-12-26 01:14:42