uva 11796 Dog Distance （几何+模拟）

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

　　一道几何模拟题。题意是，两只狗分别在两条折线上运动，已知运动轨迹，以及他们是同时到达折线的末端的。问题是，要求他们运动过程中最大距离与最小距离的差是多少。

　　表示我十分欣赏这题的做法。估计是我太饿了的缘故，居然看了天都没有想到可以将同时运动转化为相对运动，然后用静态的方法来解决这个动态的问题。如果假设A是不动的，那么这时候，我们只要搞到相对位移就可以解决问题了。

代码如下：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <set>
#include <vector>
#include <iostream>
#include <algorithm>

using namespace std;

#define REP(i, n) for (int i = 0; i < (n); i++)

struct Point {
    double x, y;
    Point() {}
    Point(double x, double y) : x(x), y(y) {}
} ;
template<class T> T sqr(T x) { return x * x;}
double ptDis(Point a, Point b) { return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));}

// basic calculations
typedef Point Vec;
Vec operator + (Vec a, Vec b) { return Vec(a.x + b.x, a.y + b.y);}
Vec operator - (Vec a, Vec b) { return Vec(a.x - b.x, a.y - b.y);}
Vec operator * (Vec a, double p) { return Vec(a.x * p, a.y * p);}
Vec operator / (Vec a, double p) { return Vec(a.x / p, a.y / p);}

const double eps = 1e-8;
int sgn(double x) { return fabs(x) < eps ? 0 : (x < 0 ? -1 : 1);}
bool operator < (Point a, Point b) { return a.x < b.x || (a.x == b.x && a.y < b.y);}
bool operator == (Point a, Point b) { return sgn(a.x - b.x) == 0 && sgn(a.y - b.y) == 0;}

double dotDet(Vec a, Vec b) { return a.x * b.x + a.y * b.y;}
double vecLen(Vec x) { return sqrt(sqr(x.x) + sqr(x.y));}
double angle(Vec a, Vec b) { return acos(dotDet(a, b) / vecLen(a) / vecLen(b));}
double crossDet(Vec a, Vec b) { return a.x * b.y - a.y * b.x;}
double triArea(Point a, Point b, Point c) { return fabs(crossDet(b - a, c - a));}
Vec rotate(Vec x, double rad) { return Vec(x.x * cos(rad) - x.y * sin(rad), x.x * sin(rad) + x.y * cos(rad));}
Vec normal(Vec x) {
    double len = vecLen(x);
    return Vec(- x.y / len, x.x / len);
}

struct Line {
    Point s, t;
    Line() {}
    Line(Point s, Point t) : s(s), t(t) {}
} ;
typedef Line Seg;

bool onSeg(Point x, Point a, Point b) { return sgn(crossDet(a - x, b - x)) == 0 && sgn(dotDet(a - x, b - x)) < 0;}
bool onSeg(Point x, Seg s) { return onSeg(x, s.s, s.t);}
// 0 : not intersect
// 1 : proper intersect
// 2 : improper intersect
int segIntersect(Point a, Point c, Point b, Point d) {
    Vec v1 = b - a, v2 = c - b, v3 = d - c, v4 = a - d;
    int a_bc = sgn(crossDet(v1, v2));
    int b_cd = sgn(crossDet(v2, v3));
    int c_da = sgn(crossDet(v3, v4));
    int d_ab = sgn(crossDet(v4, v1));
    if (a_bc * c_da > 0 && b_cd * d_ab > 0) return 1;
    if (onSeg(b, a, c) && c_da) return 2;
    if (onSeg(c, b, d) && d_ab) return 2;
    if (onSeg(d, c, a) && a_bc) return 2;
    if (onSeg(a, d, b) && b_cd) return 2;
    return 0;
}
int segIntersect(Seg a, Seg b) { return segIntersect(a.s, a.t, b.s, b.t);}

// point of the intersection of 2 lines
Point lineIntersect(Point P, Vec v, Point Q, Vec w) {
    Vec u = P - Q;
    double t = crossDet(w, u) / crossDet(v, w);
    return P + v * t;
}
Point lineIntersect(Line a, Line b) { return lineIntersect(a.s, a.t - a.s, b.s, b.t - b.s);}

// directed distance
double pt2Line(Point x, Point a, Point b) {
    Vec v1 = b - a, v2 = x - a;
    return crossDet(v1, v2) / vecLen(v1);
}
double pt2Line(Point x, Line L) { return pt2Line(x, L.s, L.t);}

double pt2Seg(Point x, Point a, Point b) {
    if (a == b) return vecLen(x - a);
    Vec v1 = b - a, v2 = x - a, v3 = x - b;
    if (sgn(dotDet(v1, v2)) < 0) return vecLen(v2);
    if (sgn(dotDet(v1, v3)) > 0) return vecLen(v3);
    return fabs(crossDet(v1, v2)) / vecLen(v1);
}
double pt2Seg(Point x, Seg s) { return pt2Seg(x, s.s, s.t);}

struct Poly {
    vector<Point> pt;
    Poly() {}
    Poly(vector<Point> pt) : pt(pt) {}
    double area() {
        double ret = 0.0;
        int sz = pt.size();
        for (int i = 1; i < sz; i++) {
            ret += crossDet(pt[i], pt[i - 1]);
        }
        return fabs(ret / 2.0);
    }
} ;

/****************** template above *******************/

const int N = 55;
Point A[N], B[N];

int main() {
//    freopen("in", "r", stdin);
    int a, b, T;
    cin >> T;
    for (int cas = 1; cas <= T; cas++) {
        cin >> a >> b;
        for (int i = 0; i < a; i++) cin >> A[i].x >> A[i].y;
        for (int i = 0; i < b; i++) cin >> B[i].x >> B[i].y;
        double lenA = 0.0, lenB = 0.0;
        for (int i = 1; i < a; i++) lenA += ptDis(A[i - 1], A[i]);
        for (int i = 1; i < b; i++) lenB += ptDis(B[i - 1], B[i]);
        Point posA = A[0], posB = B[0];
        double mn = 1e100, mx = 0.0;
        int idA = 1, idB = 1;
        while (idA < a && idB < b) {
            double disA = ptDis(posA, A[idA]);
            double disB = ptDis(posB, B[idB]);
            double minTime = min(disA / lenA, disB / lenB);
            Vec dirA = (A[idA] - posA) * minTime * lenA / disA; // displacement of A
            Vec dirB = (B[idB] - posB) * minTime * lenB / disB; // displacement of B
            mn = min(mn, pt2Seg(posA, posB, posB + dirB - dirA)); // see A as static, so it is just that B is moving
            mx = max(mx, max(ptDis(posA, posB), ptDis(posA, posB + dirB - dirA)));
            posA = posA + dirA;
            posB = posB + dirB;
            if (posA == A[idA]) idA++;
            if (posB == B[idB]) idB++;
        }
        printf("Case %d: %.f\n", cas, mx - mn);
    }
    return 0;
}

——written by Lyon