UVa 11796 Dog Distance

题意：

有甲乙两条狗分别沿着一条折线奔跑，已知它们同时从各自的起点出发，同时到达各自的终点。求整个过程中两条狗的最大距离Max与最小距离Min的差值。

分析：

假设甲乙的路线都是一条线段的简单情况。运动是相对的，我们假定甲不动，乙相对甲的运动也是匀速直线运动。所以就将问题转化成了点到直线的最小和最大距离。

甲或乙每到达一个拐点所对应的时刻称作“关键点”，那么没两个关键点之间的运动都可看做上面分析的简单情况。我们只要及时更新甲乙的位置即可。

LenA和LenB分别是两条路线的总长，因为运动时间相同，不妨设二者的运动速度为LenA和LenB，这样总时间为1

//#define LOCAL
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

struct Point
{
    double x, y;
    Point(double x=0, double y=0) :x(x),y(y) {}
};
typedef Point Vector;
Point read_point(void)
{
    double x, y;
    scanf("%lf%lf", &x, &y);
    return Point(x, y);
}
const double EPS = 1e-8;
Vector operator + (Vector A, Vector B)    { return Vector(A.x + B.x, A.y + B.y); }
Vector operator - (Vector A, Vector 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); }
bool operator < (const Point& a, const Point& b)
{ return a.x < b.x || (a.x == b.x && a.y < b.y); }
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 Cross(Vector A, Vector B)
{ return A.x*B.y - A.y*B.x; }
double DistanceToSegment(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);
}
const int maxn = 60;
Point P[maxn], Q[maxn];
double Min, Max;

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

int main(void)
{
    #ifdef    LOCAL
        freopen("11796in.txt", "r", stdin);
    #endif
    
    int T, A, B;
    scanf("%d", &T);
    for(int kase = 1; kase <= T; ++kase)
    {
        int A, B;
        scanf("%d%d", &A, &B);
        for(int i = 0; i < A; ++i)    P[i] = read_point();
        for(int i = 0; i < B; ++i)    Q[i] = read_point();

        double LenA = 0.0, LenB = 0.0;
        for(int i = 0; i < A-1; ++i)    LenA += Length(P[i+1] - P[i]);
        for(int i = 0; i < B-1; ++i)    LenB += Length(Q[i+1] - Q[i]);

        int Sa = 0, Sb = 0;        //甲乙当前端点的编号
        Point Pa = P[0], Pb = Q[0];
        Min = 1e9, Max = -1e9;
        while(Sa < A-1 && Sb < B-1)
        {
            double La = Length(P[Sa+1] - Pa);    //甲乙分别到下一拐点的距离
            double Lb = Length(Q[Sb+1] - Pb);
            double T = min(La / LenA, Lb / LenB);
            Vector Va = (P[Sa+1] - Pa) / La * T * LenA;
            Vector Vb = (Q[Sb+1] - Pb) / Lb * T * LenB;
            update(Pa, Pb, Pb + Vb - Va);
            Pa = Pa + Va;
            Pb = Pb + Vb;
            if(Pa == P[Sa+1])    Sa++;
            if(Pb == Q[Sb+1])    Sb++;
        }

        printf("Case %d: %.0lf
", kase, Max - Min);
    }
    return 0;
}