Uva 11796 Dog Distance

题意：

见白书...

思路：

我们只要按照相对运动来处理就好了，不过这里一定要理解，相对运动是通过向量来解决的，如果单纯的依靠速度来决定他的走向的话，只有两条线段平行的时候才可以。然后就是模拟这个相对运动的过程，看谁先到达拐点然后处理。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define ll long long
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d
", x)
#define lowbit(x)   (x)&(-x)
#define keyTree (chd[chd[root][1]][0])
#define Read()  freopen("din.txt", "r", stdin)
#define Write() freopen("dout.txt", "w", stdout);


#define M 100
#define N 307

using namespace std;

int dx[4]={-1,1,0,0};
int dy[4]={0,0,-1,1};//懈芯芯斜胁小褋褉

const int inf = 0x7f7f7f7f;
const int mod = 1000000007;
const double eps = 1e-8;

struct Point
{
    double x,y;
    Point(double tx = 0,double ty = 0) : x(tx),y(ty){}
};
typedef Point Vtor;
//向量的加减乘除
Vtor operator + (Vtor A,Vtor B) { return Vtor(A.x + B.x,A.y + B.y); }
Vtor operator - (Point A,Point B) { return Vtor(A.x - B.x,A.y - B.y); }
Vtor operator * (Vtor A,double p) { return Vtor(A.x*p,A.y*p); }
Vtor operator / (Vtor A,double p) { return Vtor(A.x/p,A.y/p); }
bool operator < (Point A,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 == (Point A,Point B) {return dcmp(A.x - B.x) == 0 && dcmp(A.y - B.y) == 0; }
//向量的点积，长度，夹角
double Dot(Vtor A,Vtor B) { return A.x*B.x + A.y*B.y; }
double Length(Vtor A) { return sqrt(Dot(A,A)); }
double Angle(Vtor A,Vtor B) { return acos(Dot(A,B)/Length(A)/Length(B)); }
//叉积，三角形面积
double Cross(Vtor A,Vtor B) { return A.x*B.y - A.y*B.x; }
double Area2(Point A,Point B,Point C) { return Cross(B - A,C - A); }
//向量的旋转,求向量的单位法线（即左转90度，然后长度归一）
Vtor Rotate(Vtor A,double rad){ return Vtor(A.x*cos(rad) - A.y*sin(rad),A.x*sin(rad) + A.y*cos(rad)); }
Vtor Normal(Vtor A)
{
    double L = Length(A);
    return Vtor(-A.y/L, A.x/L);
}
//直线的交点
Point GetLineIntersection(Point P,Vtor v,Point Q,Vtor w)
{
    Vtor u = P - Q;
    double t = Cross(w,u)/Cross(v,w);
    return P + v*t;
}
//点到直线的距离
double DistanceToLine(Point P,Point A,Point B)
{
    Vtor v1 = B - A;
    return Cross(P,v1)/Length(v1);
}
//点到线段的距离
double DistanceToSegment(Point P,Point A,Point B)
{
    if (A == B) return Length(P - A);
    Vtor 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);
}
//点到直线的映射
Point GetLineProjection(Point P,Point A,Point B)
{
    Vtor v = B - A;
    return A + v*Dot(v,P - A)/Dot(v,v);
}

//判断线段是否规范相交
bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2)
{
    double c1 = Cross(a2 - a1,b1 - a1), c2 = Cross(a2 - a1,b2 - a1),
           c3 = Cross(b2 - b1,a1 - b1), c4 = Cross(b2 - b1,a2 - b1);
    return dcmp(c1)*dcmp(c2) < 0 && dcmp(c3)*dcmp(c4) < 0;
}
//判断点是否在一条线段上
bool OnSegment(Point P,Point a1,Point a2)
{
    return dcmp(Cross(a1 - P,a2 - P)) == 0 && dcmp(Dot(a1 - P,a2 - P)) < 0;
}
//多边形面积
double PolgonArea(Point *p,int n)
{
    double area = 0;
    for (int i = 1; i < n - 1; ++i)
    area += Cross(p[i] - p[0],p[i + 1] - p[0]);
    return area/2;
}

Point P[N],Q[N];
double Ma,Mi;
void update(Point a,Point b,Point c)
{
    Mi = min(Mi,DistanceToSegment(a,b,c));
    Ma = max(Ma,max(Length(b - a),Length(c - a)));
}
int n,m;

int main()
{
//    Read();
    int T;
    int cas = 1;

    scanf("%d",&T);
    while (T--)
    {
        scanf("%d%d",&n,&m);
        Ma = -inf,  Mi = inf;
        for (int i = 0; i < n; ++i) scanf("%lf%lf",&P[i].x,&P[i].y);
        for (int i = 0; i < m; ++i) scanf("%lf%lf",&Q[i].x,&Q[i].y);
        
        //求出总路程，可以当做速度
        double La = 0,Lb = 0;
        for (int i = 0; i < n - 1; ++i) La += Length(P[i + 1] - P[i]);
        for (int i = 0; i < m - 1; ++i) Lb += Length(Q[i + 1] - Q[i]);

        int sa = 1,sb = 1;
        Point pa = P[0],pb = Q[0];

        while (sa < n && sb < m)
        {
            double L1 = Length(P[sa] - pa);
            double L2 = Length(Q[sb] - pb);
            double t = min(L1/La,L2/Lb);//求出到达下一个拐点的时间
            //求出位移量，注意这里是位移量。
            Vtor va = (P[sa] - pa)/L1*t*La;
            Vtor vb = (Q[sb] - pb)/L2*t*Lb;
            //更新距离
            update(pa,pb,pb + vb - va);
            //更新数据
            pa = pa + va;
            pb = pb + vb;
            if (pa == P[sa]) sa++;
            if (pb == Q[sb]) sb++;
//            printf("Sa = %d Sb = %d
",sa,sb);
        }
        printf("Case %d: %.0lf
",cas++,Ma - Mi);
    }
    return 0;
}