UVa 11796 计算几何

题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2896

大概思路：A和B运动，以A（Posa）为参考点，求出B的运动轨迹（Posb -> Posb + Vb-Va）;

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn = 60;
const int maxe = 100000;
const int INF = 0x3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1.0);

struct Point{
    double x,y;
    Point(double x=0, double y=0) : x(x),y(y){ }    //构造函数
};
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);
}

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(const Vector& A, const Vector& B) { return A.x*B.y - A.y*B.x; }

///求点P到线段AB的距离，先看Q点在线段外还是内；利用点积就可以，
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);
}


Point read_point(){
    Point A;
    scanf("%lf %lf",&A.x,&A.y);
    return A;
}

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()
{
    ///freopen("E:\\acm\\input.txt","r",stdin);
    ///freopen("E:\\acm\\output.txt","w",stdout);
    int T,a,b;
    Point A[maxn],B[maxn];
    cin>>T;
    for(int t=1;t<=T;t++){
        cin>>a>>b;
        for(int i=1;i<=a;i++)  A[i] = read_point();
        for(int i=1;i<=b;i++)  B[i] = read_point();

        double lenA = 0,lenB = 0;
        for(int i=1;i<a;i++)   lenA += Length(A[i+1]-A[i]);
        for(int i=1;i<b;i++)   lenB += Length(B[i+1]-B[i]);  //这儿也能写错；复制就是不好。

        int Sa = 1, Sb = 1;
        Point Posa = A[1], Posb = B[1];  // 现在甲乙所在的位置；
        Min =  1e9, Max = -1e9;
        while(Sa < a && Sb < b){
            double  La = Length(A[Sa+1]-Posa);
            double  Lb = Length(B[Sb+1]-Posb);
            double  T = min(La/lenA,Lb/lenB);  // 取合适的单位，可以让甲和乙的速度分别是LenA和LenB
            Vector  Va = (A[Sa+1]-Posa)/La * T * lenA;  //A在这次比较下的位移向量；
            Vector  Vb = (B[Sb+1]-Posb)/Lb * T * lenB;
            update(Posa,Posb,Posb+Vb-Va);  //Vb-Va是A在相对静止时，B的位移向量；
            Posa = Posa + Va;
            Posb = Posb + Vb;
            if(Posa == A[Sa+1])  Sa++;
            if(Posb == B[Sb+1])  Sb++;
        }
        printf("Case %d: %.0lf\n",t,Max-Min);
    }

    return 0;
}