学步园

大意：甲和乙分别沿着一条直线奔跑。两只狗的速度未知，但已知他们同时出发，同时到达，求它们在奔跑过程中的最远距离与最近距离之差。

思路：由于运动是相对的，因此可以求出甲和乙的相对位移量，认为甲不动，乙静止沿着直线走，因此问题转换为求点到线段的最小或者最大距离。然后模拟整个过程即可。

#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;
const double 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;

struct Line
{
	Point P;
	Vector v;
	double ang;
	Line() {}
	Line(Point P, Vector v) : P(P), v(v) { ang = atan2(v.y, v.x); }
	bool operator < (const Line& L) const
	{
		return ang < L.ang;
	}
};

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 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 fabs(Cross(B-A, C-A)) / 2; }

Vector Rotate(Vector A, double 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)
{
	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;
}

bool OnSegment(Point p, Point a1, Point a2)
{
	return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 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/2;
}

double PointDistanceToLine(Point P, Point A, Point B)
{
	Vector v1 = B-A, v2 = P-A;
	return fabs(Cross(v1, v2)) / Length(v1);
}

double PointDistanceToSegment(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);
	else if(dcmp(Dot(v1, v3) > 0)) return Length(v3);
	else return fabs(Cross(v1, v2)) / Length(v1);
}

int isPointInPolygon(Point p, Point *poly, int n)
{
	int wn = 0;
	for(int i = 0; i < n; i++)
	{
		const Point& p1 = poly[i], p2 = poly[(i+1)%n];
		if(p == p1 || p == p2 || OnSegment(p, p1, p2)) return -1;
		int k = dcmp(Cross(p2-p1, p-p1));
		int d1 = dcmp(p1.y - p.y);
		int d2 = dcmp(p2.y - p.y);
		if(k > 0 && d1 <= 0 && d2 > 0) wn++;
		if(k < 0 && d2 <= 0 && d1 > 0) wn--;
	}
	if(wn != 0) return 1;
	return 0;
}

int ConvexHull(Point *p, int n, Point *ch) //凸包
{
	sort(p, p+n);
	n = unique(p, p+n) - p; //去重
	int m = 0;
	for(int i = 0; i < n; i++)
	{
		while(m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
		ch[m++] = p[i];
	}
	int k = m;
	for(int i = n-2; i >= 0; i--)
	{
		while(m > k && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
		ch[m++] = p[i];
	}
	if(n > 1) m--;
	return m;
}

Vector Normal(Vector A)
{
	double L = Length(A);
	return Vector (-A.y/L, A.x/L);
}

double Dist2(Point p1, Point p2)
{
	return (p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y);
}

// 返回点集直径的平方
double RotatingCalipers(Point *P, int n) //旋转卡壳 
{
  if(n == 1) return 0;
  if(n == 2) return Dist2(P[0], P[1]);
  P[n] = P[0]; //避免取模
  double ans = 0;
  for(int u = 0, v = 1; u < n; u++)
  {
	//一条直线贴住边p[u]-p[u+1]
    for(;;)
	{
	  // 当Area(p[u], p[u+1], p[v+1]) <= Area(p[u], p[u+1], p[v])时停止旋转
      // 即Cross(p[u+1]-p[u], p[v+1]-p[u]) - Cross(p[u+1]-p[u], p[v]-p[u]) <= 0
      // 根据Cross(A,B) - Cross(A,C) = Cross(A,B-C)
      // 化简得Cross(p[u+1]-p[u], p[v+1]-p[v]) <= 0
      double diff = Cross(P[u+1]-P[u], P[v+1]-P[v]);
      if(diff <= 0)
      {
        ans = max(ans, Dist2(P[u], P[v])); // u和v是对踵点
        if(diff == 0) ans = max(ans, Dist2(P[u], P[v+1])); // diff == 0时u和v+1也是对踵点
        break; //即边平行 
      }
      v = (v + 1) % n;
    }
  }
  return ans;
}

bool OnLeft(Line L, Point p)
{
	return Cross(L.v, p-L.P) > 0;
}

Point GetIntersection(Line a, Line b)
{
	Vector u = a.P-b.P;
	double t = Cross(b.v, u) / Cross(a.v, b.v);
	return a.P+a.v*t;
}

int HalfPlaneIntersection(Line* L, int n, Point* poly)
{
	sort(L, L+n);
	int first, last;
	Point *p = new Point[n];
	Line* q = new Line[n];
	q[first=last=0] = L[0];
	for(int i = 1; i < n; i++)
	{
		while(first < last && !OnLeft(L[i], p[last-1])) last--;
		while(first < last && !OnLeft(L[i], p[first])) first++;
		q[++last] = L[i];
		if(fabs(Cross(q[last].v, q[last-1].v)) < eps)
		{
			last--;
			if(OnLeft(q[last], L[i].P)) q[last] = L[i];
		}
		if(first < last) p[last-1] = GetIntersection(q[last-1], q[last]);
	}
	while(first < last && !OnLeft(q[first], p[last-1])) last--;
	if(last - first <= 1) return 0;
	p[last] = GetIntersection(q[last], q[first]);
	
	int m = 0;
	for(int i = first; i <= last; i++) poly[m++] = p[i];
	return m;
}

Point read_point()
{
	Point A;
	scanf("%lf%lf", &A.x, &A.y);
	return A;
}

int n, m;

double Max, Min;

const int maxn = 60;
const double INF = 0x3f3f3f3f;

Point P[maxn], Q[maxn];

void init()
{
	Max = -INF;
	Min = INF;
}

void read_case()
{
	init();
	scanf("%d%d", &n, &m);
	for(int i = 0; i < n; i++) P[i] = read_point();
	for(int i = 0; i < m; i++) Q[i] = read_point();
}

void update(Point P, Point A, Point B)
{
	Min = min(Min, PointDistanceToSegment(P, A, B));
	Max = max(Max, max(Length(P-A), Length(P-B)));
}

void solve()
{
	read_case();
	int Sa = 0, Sb = 0;
	double LenA = 0, LenB = 0;
	for(int i = 0; i < n-1; i++) LenA += Length(P[i+1]-P[i]);
	for(int i = 0; i < m-1; i++) LenB += Length(Q[i+1]-Q[i]);
	Point Pa = P[0], Pb = Q[0];
	while(Sa < n-1 && Sb < m-1)
	{
		double La = Length(P[Sa+1]-Pa);
		double Lb = Length(Q[Sb+1]-Pb);
		double T = min(La/LenA, Lb/LenB);
		Vector Va = ((P[Sa+1]-Pa)/La)*T*LenA; //位移向量
		Vector Vb = ((Q[Sb+1]-Pb)/Lb)*T*LenB; //位移向量
		update(Pa, Pb, Pb+Vb-Va); //相对位移 
		Pa = Pa + Va;
		Pb = Pb + Vb;
		if(Pa == P[Sa+1]) Sa++;
		if(Pb == Q[Sb+1]) Sb++;
	}
	printf("%.0lf\n", Max-Min);
}

int main()
{
	int T, times = 0;
	scanf("%d", &T);
	while(T--)
	{
		printf("Case %d: ", ++times);
		solve();
	}
	return 0;
}