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UVa 11796 Dog Distance

时间:2014-10-13 22:42:17      阅读:234      评论:0      收藏:0      [点我收藏+]

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题意:

有甲乙两条狗分别沿着一条折线奔跑,已知它们同时从各自的起点出发,同时到达各自的终点。求整个过程中两条狗的最大距离Max与最小距离Min的差值。

分析:

假设甲乙的路线都是一条线段的简单情况。运动是相对的,我们假定甲不动,乙相对甲的运动也是匀速直线运动。所以就将问题转化成了点到直线的最小和最大距离。

甲或乙每到达一个拐点所对应的时刻称作“关键点”,那么没两个关键点之间的运动都可看做上面分析的简单情况。我们只要及时更新甲乙的位置即可。

LenA和LenB分别是两条路线的总长,因为运动时间相同,不妨设二者的运动速度为LenA和LenB,这样总时间为1

 

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 1 //#define LOCAL
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <cmath>
 6 using namespace std;
 7 
 8 struct Point
 9 {
10     double x, y;
11     Point(double x=0, double y=0) :x(x),y(y) {}
12 };
13 typedef Point Vector;
14 Point read_point(void)
15 {
16     double x, y;
17     scanf("%lf%lf", &x, &y);
18     return Point(x, y);
19 }
20 const double EPS = 1e-8;
21 Vector operator + (Vector A, Vector B)    { return Vector(A.x + B.x, A.y + B.y); }
22 Vector operator - (Vector A, Vector B)    { return Vector(A.x - B.x, A.y - B.y); }
23 Vector operator * (Vector A, double p)    { return Vector(A.x*p, A.y*p); }
24 Vector operator / (Vector A, double p)    { return Vector(A.x/p, A.y/p); }
25 bool operator < (const Point& a, const Point& b)
26 { return a.x < b.x || (a.x == b.x && a.y < b.y); }
27 int dcmp(double x)
28 { if(fabs(x) < EPS) return 0; else return x < 0 ? -1 : 1; }
29 bool operator == (const Point& a, const Point& b)
30 { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; }
31 double Dot(Vector A, Vector B)
32 { return A.x*B.x + A.y*B.y; }
33 double Length(Vector A)    { return sqrt(Dot(A, A)); }
34 
35 double Cross(Vector A, Vector B)
36 { return A.x*B.y - A.y*B.x; }
37 double DistanceToSegment(Point P, Point A, Point B)
38 {
39     if(A == B)    return Length(P - A);
40     Vector v1 = B - A, v2 = P - A, v3 = P - B;
41     if(dcmp(Dot(v1, v2)) < 0)    return Length(v2);
42     else if(dcmp(Dot(v1, v3)) > 0)    return Length(v3);
43     else return fabs(Cross(v1, v2)) / Length(v1);
44 }
45 const int maxn = 60;
46 Point P[maxn], Q[maxn];
47 double Min, Max;
48 
49 void update(Point P, Point A, Point B)
50 {
51     Min = min(Min, DistanceToSegment(P, A, B));
52     Max = max(Max, Length(P-A));
53     Max = max(Max, Length(P-B));
54 }
55 
56 int main(void)
57 {
58     #ifdef    LOCAL
59         freopen("11796in.txt", "r", stdin);
60     #endif
61     
62     int T, A, B;
63     scanf("%d", &T);
64     for(int kase = 1; kase <= T; ++kase)
65     {
66         int A, B;
67         scanf("%d%d", &A, &B);
68         for(int i = 0; i < A; ++i)    P[i] = read_point();
69         for(int i = 0; i < B; ++i)    Q[i] = read_point();
70 
71         double LenA = 0.0, LenB = 0.0;
72         for(int i = 0; i < A-1; ++i)    LenA += Length(P[i+1] - P[i]);
73         for(int i = 0; i < B-1; ++i)    LenB += Length(Q[i+1] - Q[i]);
74 
75         int Sa = 0, Sb = 0;        //甲乙当前端点的编号
76         Point Pa = P[0], Pb = Q[0];
77         Min = 1e9, Max = -1e9;
78         while(Sa < A-1 && Sb < B-1)
79         {
80             double La = Length(P[Sa+1] - Pa);    //甲乙分别到下一拐点的距离
81             double Lb = Length(Q[Sb+1] - Pb);
82             double T = min(La / LenA, Lb / LenB);
83             Vector Va = (P[Sa+1] - Pa) / La * T * LenA;
84             Vector Vb = (Q[Sb+1] - Pb) / Lb * T * LenB;
85             update(Pa, Pb, Pb + Vb - Va);
86             Pa = Pa + Va;
87             Pb = Pb + Vb;
88             if(Pa == P[Sa+1])    Sa++;
89             if(Pb == Q[Sb+1])    Sb++;
90         }
91 
92         printf("Case %d: %.0lf\n", kase, Max - Min);
93     }
94     return 0;
95 }
代码君

 

UVa 11796 Dog Distance

标签:style   blog   http   color   io   os   ar   for   sp   

原文地址:http://www.cnblogs.com/AOQNRMGYXLMV/p/4023078.html

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