UV 11796 dog distance

Question:

Two dogs, A and B, run along a broken line. It is known that they start from their respective starting points and reach their respective ending points at the same time. Evaluate the difference between the maximum distance between the two dogs and the minimum distance Min.

Analysis:

Assume that the routes of A and B are all a line segment. The motion is relative. We assume that a does not move, and B is moving at a constant speed and straight line. Therefore, the problem is transformed into the minimum and maximum distance from a point to a straight line.

When Party A or Party B reach an inflection point, it is called a "Key Point". The movement between the two key points can be seen as a simple situation analyzed above. We only need to update the location of Party A and Party B in time.

Lena and lenb are the total lengths of the two routes. Because the motion time is the same, you can set the motion speed of the two routes to Lena and lenb. In this way, the total time is 1.

 

 1 //#define LOCAL 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 using namespace std; 7  8 struct Point 9 {10     double x, y;11     Point(double x=0, double y=0) :x(x),y(y) {}12 };13 typedef Point Vector;14 Point read_point(void)15 {16     double x, y;17     scanf("%lf%lf", &x, &y);18     return Point(x, y);19 }20 const double EPS = 1e-8;21 Vector operator + (Vector A, Vector B)    { return Vector(A.x + B.x, A.y + B.y); }22 Vector operator - (Vector A, Vector B)    { return Vector(A.x - B.x, A.y - B.y); }23 Vector operator * (Vector A, double p)    { return Vector(A.x*p, A.y*p); }24 Vector operator / (Vector A, double p)    { return Vector(A.x/p, A.y/p); }25 bool operator < (const Point& a, const Point& b)26 { return a.x < b.x || (a.x == b.x && a.y < b.y); }27 int dcmp(double x)28 { if(fabs(x) < EPS) return 0; else return x < 0 ? -1 : 1; }29 bool operator == (const Point& a, const Point& b)30 { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; }31 double Dot(Vector A, Vector B)32 { return A.x*B.x + A.y*B.y; }33 double Length(Vector A)    { return sqrt(Dot(A, A)); }34 35 double Cross(Vector A, Vector B)36 { return A.x*B.y - A.y*B.x; }37 double DistanceToSegment(Point P, Point A, Point B)38 {39     if(A == B)    return Length(P - A);40     Vector v1 = B - A, v2 = P - A, v3 = P - B;41     if(dcmp(Dot(v1, v2)) < 0)    return Length(v2);42     else if(dcmp(Dot(v1, v3)) > 0)    return Length(v3);43     else return fabs(Cross(v1, v2)) / Length(v1);44 }45 const int maxn = 60;46 Point P[maxn], Q[maxn];47 double Min, Max;48 49 void update(Point P, Point A, Point B)50 {51     Min = min(Min, DistanceToSegment(P, A, B));52     Max = max(Max, Length(P-A));53     Max = max(Max, Length(P-B));54 }55 56 int main(void)57 {58     #ifdef    LOCAL59         freopen("11796in.txt", "r", stdin);60     #endif61     62     int T, A, B;63     scanf("%d", &T);64     for(int kase = 1; kase <= T; ++kase)65     {66         int A, B;67         scanf("%d%d", &A, &B);68         for(int i = 0; i < A; ++i)    P[i] = read_point();69         for(int i = 0; i < B; ++i)    Q[i] = read_point();70 71         double LenA = 0.0, LenB = 0.0;72         for(int i = 0; i < A-1; ++i)    LenA += Length(P[i+1] - P[i]);73         for(int i = 0; i < B-1; ++i)    LenB += Length(Q[i+1] - Q[i]);74 75         int Sa = 0, Sb = 0;        //甲乙当前端点的编号76         Point Pa = P[0], Pb = Q[0];77         Min = 1e9, Max = -1e9;78         while(Sa < A-1 && Sb < B-1)79         {80             double La = Length(P[Sa+1] - Pa);    //甲乙分别到下一拐点的距离81             double Lb = Length(Q[Sb+1] - Pb);82             double T = min(La / LenA, Lb / LenB);83             Vector Va = (P[Sa+1] - Pa) / La * T * LenA;84             Vector Vb = (Q[Sb+1] - Pb) / Lb * T * LenB;85             update(Pa, Pb, Pb + Vb - Va);86             Pa = Pa + Va;87             Pb = Pb + Vb;88             if(Pa == P[Sa+1])    Sa++;89             if(Pb == Q[Sb+1])    Sb++;90         }91 92         printf("Case %d: %.0lf\n", kase, Max - Min);93     }94     return 0;95 }

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UV 11796 dog distance