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LA 2218 (半平面交) Triathlon

题意:

有n个选手,铁人三项有连续的三段,对于每段场地选手i分别以vi, ui 和 wi匀速通过。

对于每个选手,问能否通过调整每种赛道的长度使得他成为冠军(不能并列)。

分析:

粗一看,这不像一道计算几何的题目。

假设赛道总长度是1,第一段长x,第二段长y,第三段则是1-x-y

那么可以计算出每个选手完成比赛的时间Ti

对于选手i,若要成为冠军则有Ti < Tj (i ≠ j)

于是就有n-1个不等式,每个不等式都代表一个半平面。

在加上x>0, y>0, 1-x-y>0 这三个半平面一共有n+2个半平面。如果这些半平面交非空,则选手i可以成为冠军。

最终,还是转化成了半平面交的问题。

 

细节:

  • 对于半平面 ax+by+c > 0 所对应的向量(b, -a)是和系数的正负没有关系的,可以自己试验下。开始我纠结了好长时间
  • if(fabs(a) > fabs(b))    P = Point(-c/a, 0)else P = Point(0, -c/b);

    对于这段代码不是太清楚它的含义,因为不管怎样P都是在ax+by+c = 0 这条直线上的。我猜可能是减小浮点运算的误差吧?

 

  1 //#define LOCAL  2 #include <cstdio>  3 #include <cstring>  4 #include <algorithm>  5 #include <cmath>  6 #include <vector>  7 using namespace std;  8   9 const double eps = 1e-6; 10 const int maxn = 100 + 10; 11 int v[maxn], u[maxn], w[maxn]; 12  13 struct Point 14 { 15     double x, y; 16     Point(double x=0, double y=0):x(x), y(y) {} 17 }; 18 typedef Point Vector; 19 Point operator + (Point a, Point b) { return Point(a.x+b.x, a.y+b.y); } 20 Point operator - (Point a, Point b) { return Point(a.x-b.x, a.y-b.y); } 21 Vector operator * (Vector a, double p) { return Vector(a.x*p, a.y*p); } 22 Vector operator / (Vector a, double p) { return Vector(a.x/p, a.y/p); } 23 bool operator < (const Point& a, const Point& b) 24 { return a.x < b.x || (a.x == b.x && a.y < b.y); } 25 bool operator == (const Point& a, const Point& b) 26 { return a.x == b.x && a.y == b.y; } 27 double Dot(const Vector& a, const Vector& b) { return a.x*b.x + a.y*b.y; } 28 double Cross(const Vector& a, const Vector& b) { return a.x*b.y - a.y*b.x; } 29 double Length(const Vector& a) { return sqrt(Dot(a, a)); } 30 Vector Normal(const Vector& a) 31 { 32     double l = Length(a); 33     return Vector(-a.y/l, a.x); 34 } 35  36 double PolygonArea(const vector<Point>& p) 37 { 38     int n = p.size(); 39     double ans = 0.0; 40     for(int i = 1; i < n-1; ++i) 41         ans += Cross(p[i]-p[0], p[i+1]-p[0]);     42     return ans/2; 43 } 44  45 struct Line 46 { 47     Point P; 48     Vector v; 49     double ang; 50     Line() {} 51     Line(Point p, Vector v):P(p), v(v) { ang = atan2(v.y, v.x); } 52     bool operator < (const Line& L) const 53     { 54         return ang < L.ang; 55     } 56 }; 57  58 bool OnLeft(const Line& L, Point p) 59 { 60     return Cross(L.v, p-L.P) > 0; 61 } 62  63 Point GetLineIntersevtion(const Line& a, const Line& b) 64 { 65     Vector u = a.P - b.P; 66     double t = Cross(b.v, u) / Cross(a.v, b.v); 67     return a.P + a.v*t; 68 } 69  70 vector<Point> HalfplaneIntersection(vector<Line> L) 71 { 72     int n = L.size(); 73     sort(L.begin(), L.end()); 74  75     int first, last; 76     vector<Point> p(n); 77     vector<Line> q(n); 78     vector<Point> ans; 79  80     q[first=last=0] = L[0]; 81     for(int i = 1; i < n; ++i) 82     { 83         while(first < last && !OnLeft(L[i], p[last-1])) last--; 84         while(first < last && !OnLeft(L[i], p[first])) first++; 85         q[++last] = L[i]; 86         if(fabs(Cross(q[last].v, q[last-1].v)) < eps) 87         { 88             last--; 89             if(OnLeft(q[last], L[i].P)) q[last] = L[i]; 90         } 91         if(first < last) p[last-1] = GetLineIntersevtion(q[last-1], q[last]); 92     } 93     while(first < last && !OnLeft(q[first], p[last-1])) last--; 94     if(last - first <= 1)    return ans; 95     p[last] = GetLineIntersevtion(q[first], q[last]); 96  97     for(int i = first; i <= last; ++i)    ans.push_back(p[i]); 98     return ans; 99 }100 101 int main(void)102 {103     #ifdef    LOCAL104         freopen("2218in.txt", "r", stdin);105     #endif106     107     int n;108     while(scanf("%d", &n) == 1 && n)109     {110         for(int i = 0; i < n; ++i) scanf("%d%d%d", &v[i], &u[i], &w[i]);111         for(int i = 0; i < n; ++i)112         {113             int ok = 1;114             double k = 10000;115             vector<Line> L;116             for(int j = 0; j < n; ++j) if(j != i)117             {118                 if(v[i]<=v[j] && u[i]<=u[j] && w[i]<=w[j]) { ok = 0; break; }119                 if(v[i]>v[j] && u[i]>u[j] && w[i]>w[j]) continue;120 121                 double a = (k/v[j]-k/v[i]+k/w[i]-k/w[j]);122                 double b = (k/u[j]-k/u[i]+k/w[i]-k/w[j]);123                 double c = k/w[j]-k/w[i];124                 //L.push_back(Line(Point(0, -c/b), Vector(b, -a)));125                 Point P;126                 Vector V(b, -a);127                 if(fabs(a) > fabs(b))    P = Point(-c/a, 0);128                 else P = Point(0, -c/b);129                 L.push_back(Line(P, V));130             }131             if(ok)132             {133                 L.push_back(Line(Point(0, 0), Vector(0, -1)));134                 L.push_back(Line(Point(0, 0), Vector(1, 0)));135                 L.push_back(Line(Point(0, 1), Vector(-1, 1)));136                 vector<Point> Poly = HalfplaneIntersection(L);137                 if(Poly.empty()) ok = 0;138             }139             if(ok) puts("Yes"); else puts("No");140         }141     }142 143     return 0;144 }
代码君

 

LA 2218 (半平面交) Triathlon