UVA 11796

題意：給定兩條狗的行走路線，一直兩條狗同時出發同時到達，問路途中的最遠和最近距離

思路：把每一段路程，當成相對運動，這樣就是一只狗靜止，一只在動，最小值相當於求點到線段距離，最大值為點到線段兩端距離

代碼：

#include 
#include 
#include 
#include 
using namespace std;

struct Point {
    double x, y;
    Point() {}
    Point(double x, double y) {
        this->x = x;
        this->y = y;
    }
    void read() {
        scanf("%lf%lf", &x, &y);
    }
};

typedef Point Vector;

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);
}

const double eps = 1e-8;

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;} //點積
double Length(Vector A) {return sqrt(Dot(A, A));} //向量的模
double Angle(Vector A, Vector B) {return acos(Dot(A, B) / Length(A) / Length(B));} //向量夾角
double Cross(Vector A, Vector B) {return A.x * B.y - A.y * B.x;} //叉積

double dist(Point a, Point b) {
    Vector tmp = a - b;
    return sqrt(tmp.x * tmp.x + tmp.y * tmp.y);
}

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 N = 55;
int T, n, m;
Point a[N], b[N];
double Min, Max;

void gao(Point A, Point B, Point C) {
    Min = min(Min, DistanceToSegment(A, B, C));
    Max = max(Max, dist(A, B));
    Max = max(Max, dist(A, C));
}

int main() {
    int cas = 0;
    scanf("%d", &T);
    while (T--) {
        scanf("%d%d", &n, &m);
        for (int i = 0; i < n; i++) a[i].read();
        for (int i = 0; i < m; i++) b[i].read();
        double va = 0, vb = 0;
        Point pa = a[0], pb = b[0];
        for (int i = 0; i < n - 1; i++) va += dist(a[i + 1], a[i]);
        for (int i = 0; i < m - 1; i++) vb += dist(b[i + 1], b[i]);
        int ua = 0, ub = 0;
        Min = 1e20, Max = -1e20;
        while (ua < n - 1 && ub < m - 1) {
            double sa = dist(pa, a[ua + 1]);
            double sb = dist(pb, b[ub + 1]);
            double t = min(sa / va, sb / vb);
            Vector Va = (a[ua + 1] - pa) / sa * t * va;
            Vector Vb = (b[ub + 1] - pb) / sb * t * vb;
            gao(pa, pb, pb - Va + Vb);
            pa = pa + Va;
            pb = pb + Vb;
            if (pa == a[ua + 1]) ua++;
            if (pb == b[ub + 1]) ub++;
        }
        printf("Case %d: %.0lf\n", ++cas, Max - Min);
    }
    return 0;
}