UVA 11796

题意：  有两个狗， 按照 多边形跑，不知道两条狗的速度，但是狗是同时出发，同时到达终点的

输出两条狗的 最大相距距离 - 最小相距距离；

思路 ： 用物理的相对运动来计算， 每次只计算 两条狗的直线运动， 转折点再额外更新

LRJ 模板大法好 ！！！LRJ 模板大法好 ！！！！LRJ 模板大法好 ！！！！

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int maxn = 100;
const double eps = 1e-9;

struct Point {
    double x , y;
    Point (double x = 0, double y = 0) : x(x),y(y) {}
};

typedef Point Vector;

int dcmp (double x) { if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; }

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);     }
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;
}

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 DistanceToSegment (Point P, Point A, Point B) {   ///点到线段的距离
    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);
}

double MAX,MIN;

void UpDate(Point P, Point A, Point B)
{
    MIN = min(MIN, DistanceToSegment(P,A,B)); ///当前 和 点 到线段的最小距离
    MAX = max(Length(P-A),MAX);
    MAX = max(Length(P-B),MAX);
}

Point Pa[maxn], Pb[maxn];

int main()
{
    int t;
    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) cin >> Pa[i].x >> Pa[i].y;
        for(int i = 0; i < B; ++i) cin >> Pb[i].x >> Pb[i].y;

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

        MAX = -1e9; MIN = 1e9;
        int sa = 0, sb = 0;           ///下一个转折点
        Point Na = Pa[0], Nb = Pb[0]; /// 起始位置
        while(sa < A-1 && sb < B-1)
        {
            double La = Length(Pa[sa+1] - Na);          /// a 当前到下一个转折点的长度
            double Lb = Length(Pb[sb+1] - Nb);          /// b ...
            double t  = min(La / LenA, Lb / LenB);      /// 运动的时间
            Vector Va = (Pa[sa+1] - Na) / La * t * LenA;/// a 的位移量
            Vector Vb = (Pb[sb+1] - Nb) / Lb * t * LenB;/// b 的位移量
            UpDate(Na,Nb,Nb+Vb-Va);  /// B相对A 的运动就是 NB + Vb - Va；
            Na = Na + Va;            /// 更新 a 的当前点
            Nb = Nb + Vb;
            if(Na == Pa[sa+1]) sa++;  ///如果到了转折点， sa 下一个转折点更新
            if(Nb == Pb[sb+1]) sb++;
        }
        printf("Case %d: %.0lf\n",kase, MAX - MIN);
    }
    return 0;
}