uva 11796

题意：两仅仅狗分别沿一条折线跑，速度未知，但同一时候出发同一时候到达。分别给出两条折线的各点坐标，问两仅仅狗在跑的过程中的最远距离和近期距离差。
题解：假设两仅仅狗跑的都是线段，问题会更easy解决，能够把当中一点当做精巧，还有一个做相对运动，就能够求出这个过程的最远近期距离，然后假设是走折线，两个点先到拐点的之前的时间段就是前面说的简化过程，也就是把整个过程划分为好多个简化过程阶段。

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
const int N = 60;
struct Point {
    double x, y;
    Point(double x = 0, double y = 0): x(x), y(y) {}
}A[N], B[N];
int n, m;
double maxx, minn;

Point operator + (Point A, Point B) {
    return Point(A.x + B.x, A.y + B.y);
}

Point operator - (Point A, Point B) {
    return Point(A.x - B.x, A.y - B.y);
}

Point operator * (Point A, double p) {
    return Point(A.x * p, A.y * p);
}

Point operator / (Point A, double p) {
    return Point(A.x / p, A.y / p);
}
//计算点积的正负  负值夹角为钝角
int dcmp(double x) {
    if (fabs(x) < 1e-9)
        return 0;
    return x < 0 ?
 -1 : 1;
}

bool operator < (const Point& a, const Point& b) {
    return a.x < b.x || (a.x == b.x && a.y < b.y);
}

bool operator == (const Point& a, const Point& b) {
    return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}
//计算点积
double Dot(Point A, Point B) {
    return A.x * B.x + A.y * B.y;
}
//计算叉积，也就是数量积
double Cross(Point A, Point B) {
    return A.x * B.y - A.y * B.x;
}
//计算向量长度
double Length(Point A) {
    return sqrt(Dot(A, A));
}
//向量A旋转rad弧度，rad负值为顺时针旋转
Point Rotate(Point A, double rad) {
    return Point(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}
//得到两直线交点 
Point GetLineIntersection(Point P, Point v, Point Q, Point w) {
    Point u = P - Q;
    double t = Cross(w, u) / Cross(v, w);
    return P + v * t;
}
//点p到线段AB的距离
double DistanceToSegment(Point p, Point A, Point B) {
    if (A == B)
        return Length(p - A);
    Point AB = B - A, AP = p - A, BP = p - B;
    if (dcmp(Dot(AB, AP)) < 0)
        return Length(AP);
    else if (dcmp(Dot(AB, BP)) > 0)
        return Length(BP);
    else
        return fabs(Cross(AB, AP)) / Length(AB);
}

//推断两个线段是否有交点(不包含端点)
bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) {
    double c1 = Cross(a2 - a1, b1 - a1);
    double c2 = Cross(a2 - a1, b2 - a1);
    double c3 = Cross(b2 - b1, a1 - b1);
    double c4 = Cross(b2 - b1, a2 - b1);
    return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
}
//推断点p是否在线段a1--a2上(不包含端点)
bool OnSegment(Point p, Point a1, Point a2) {
    return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0;
}

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

int main() {
    int t, cas = 1;
    scanf("%d", &t);
    while (t--) {
        scanf("%d%d", &n, &m);
        double sum1 = 0, sum2 = 0;
        for (int i = 0; i < n; i++) {
            scanf("%lf%lf", &A[i].x, &A[i].y);
            if (i != 0)
                sum1 += Length(A[i] - A[i - 1]);
        }
        for (int i = 0; i < m; i++) {
            scanf("%lf%lf", &B[i].x, &B[i].y);
            if (i != 0)
                sum2 += Length(B[i] - B[i - 1]);
        }
        int cur1 = 0, cur2 = 0;
        Point pos1 = A[cur1], pos2 = B[cur2];
        maxx = -1e9;
        minn = 1e9;
        while (cur1 < n - 1 && cur2 < m - 1) {
            double len1 = Length(A[cur1 + 1] - pos1);
            double len2 = Length(B[cur2 + 1] - pos2);
            double T = min(len1 / sum1, len2 / sum2);
            Point v1 = (A[cur1 + 1] - pos1) / len1 * T * sum1; //甲的位移
            Point v2 = (B[cur2 + 1] - pos2) / len2 * T * sum2; //乙的位移
            update(pos1, pos2, pos2 + v2 - v1);//相对位移
            pos1 = pos1 + v1;
            pos2 = pos2 + v2;
            if (pos1 == A[cur1 + 1])
                cur1++;
            if (pos2 == B[cur2 + 1])
                cur2++;
        }
        printf("Case %d: %.0lf\n", cas++, maxx - minn);
    }
    return 0;
}