uva 2218 Triathlon

  1 #include<cstdio>
  2 #include<cmath>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<iostream>
  6 #include<memory.h>
  7 #include<cstdlib>
  8 #include<vector>
  9 #define clc(a,b) memset(a,b,sizeof(a))
 10 #define LL long long int
 11 #define up(i,x,y) for(i=x;i<=y;i++)
 12 #define w(a) while(a)
 13 const double inf=0x3f3f3f3f;
 14 const int N = 4010;
 15 const double eps = 5*1e-13;
 16 const double PI = acos(-1.0);
 17 using namespace std;
 18 
 19 struct Point
 20 {
 21     double x, y;
 22     Point(double x=0, double y=0):x(x),y(y) { }
 23 };
 24 
 25 typedef Point Vector;
 26 
 27 Vector operator + (const Vector& A, const Vector& B)
 28 {
 29     return Vector(A.x+B.x, A.y+B.y);
 30 }
 31 Vector operator - (const Point& A, const Point& B)
 32 {
 33     return Vector(A.x-B.x, A.y-B.y);
 34 }
 35 Vector operator * (const Vector& A, double p)
 36 {
 37     return Vector(A.x*p, A.y*p);
 38 }
 39 double Dot(const Vector& A, const Vector& B)
 40 {
 41     return A.x*B.x + A.y*B.y;
 42 }
 43 double Cross(const Vector& A, const Vector& B)
 44 {
 45     return A.x*B.y - A.y*B.x;
 46 }
 47 double Length(const Vector& A)
 48 {
 49     return sqrt(Dot(A, A));
 50 }
 51 Vector Normal(const Vector& A)
 52 {
 53     double L = Length(A);
 54     return Vector(-A.y/L, A.x/L);
 55 }
 56 
 57 double PolygonArea(vector<Point> p)
 58 {
 59     int n = p.size();
 60     double area = 0;
 61     for(int i = 1; i < n-1; i++)
 62         area += Cross(p[i]-p[0], p[i+1]-p[0]);
 63     return area/2;
 64 }
 65 
 66 // 有向直线。它的左边就是对应的半平面
 67 struct Line
 68 {
 69     Point P;    // 直线上任意一点
 70     Vector v;   // 方向向量
 71     double ang; // 极角，即从x正半轴旋转到向量v所需要的角（弧度）
 72     Line() {}
 73     Line(Point P, Vector v):P(P),v(v)
 74     {
 75         ang = atan2(v.y, v.x);
 76     }
 77     bool operator < (const Line& L) const
 78     {
 79         return ang < L.ang;
 80     }
 81 };
 82 
 83 // 点p在有向直线L的左边（线上不算）
 84 bool OnLeft(const Line& L, const Point& p)
 85 {
 86     return Cross(L.v, p-L.P) > 0;
 87 }
 88 
 89 // 二直线交点，假定交点惟一存在
 90 Point GetLineIntersection(const Line& a, const Line& b)
 91 {
 92     Vector u = a.P-b.P;
 93     double t = Cross(b.v, u) / Cross(a.v, b.v);
 94     return a.P+a.v*t;
 95 }
 96 
 97 const double INF = 1e8;
 98 
 99 // 半平面交主过程
100 vector<Point> HalfplaneIntersection(vector<Line> L)
101 {
102     int n = L.size();
103     sort(L.begin(), L.end()); // 按极角排序
104 
105     int first, last;         // 双端队列的第一个元素和最后一个元素的下标
106     vector<Point> p(n);      // p[i]为q[i]和q[i+1]的交点
107     vector<Line> q(n);       // 双端队列
108     vector<Point> ans;       // 结果
109 
110     q[first=last=0] = L[0];  // 双端队列初始化为只有一个半平面L[0]
111     for(int i = 1; i < n; i++)
112     {
113         while(first < last && !OnLeft(L[i], p[last-1])) last--;
114         while(first < last && !OnLeft(L[i], p[first])) first++;
115         q[++last] = L[i];
116         if(fabs(Cross(q[last].v, q[last-1].v)) < eps)   // 两向量平行且同向，取内侧的一个
117         {
118             last--;
119             if(OnLeft(q[last], L[i].P)) q[last] = L[i];
120         }
121         if(first < last) p[last-1] = GetLineIntersection(q[last-1], q[last]);
122     }
123     while(first < last && !OnLeft(q[first], p[last-1])) last--; // 删除无用平面
124     if(last - first <= 1) return ans; // 空集
125     p[last] = GetLineIntersection(q[last], q[first]); // 计算首尾两个半平面的交点
126 
127     // 从deque复制到输出中
128     for(int i = first; i <= last; i++) ans.push_back(p[i]);
129     return ans;
130 }
131 
132 const int maxn = 100 + 10;
133 int V[maxn], U[maxn], W[maxn];
134 int main()
135 {
136     int n;
137     while(scanf("%d", &n) == 1 && n)
138     {
139         for(int i = 0; i < n; i++) scanf("%d%d%d", &V[i], &U[i], &W[i]);
140         for(int i = 0; i < n; i++)
141         {
142             int ok = 1;
143             double k = 10000;
144             vector<Line> L;
145             for(int j = 0; j < n; j++) if(i != j)
146                 {
147                     if(V[i] <= V[j] && U[i] <= U[j] && W[i] <= W[j])
148                     {
149                         ok = 0;
150                         break;
151                     }
152                     if(V[i] >= V[j] && U[i] >= U[j] && W[i] >= W[j]) continue;
153                     // x/V[i]+y/U[i]+(1-x-y)/W[i] < x/V[j]+y/U[j]+(1-x-y)/W[j]
154                     // ax+by+c>0
155                     double a = (k/V[j]-k/W[j]) - (k/V[i]-k/W[i]);
156                     double b = (k/U[j]-k/W[j]) - (k/U[i]-k/W[i]);
157                     double c = k/W[j] - k/W[i];
158                     Point P;
159                     Vector v(b, -a);
160                     if(fabs(a) > fabs(b))
161                         P = Point(-c/a, 0);
162                     else
163                         P = Point(0, -c/b);
164                     L.push_back(Line(P, v));
165                 }
166             if(ok)
167             {
168                 // x>0, y>0, x+y<1 ==> -x-y+1>0
169                 L.push_back(Line(Point(0, 0), Vector(0, -1)));
170                 L.push_back(Line(Point(0, 0), Vector(1, 0)));
171                 L.push_back(Line(Point(0, 1), Vector(-1, 1)));
172                 vector<Point> poly = HalfplaneIntersection(L);
173                 if(poly.empty()) ok = 0;
174             }
175             if(ok) printf("Yes\n");
176             else printf("No\n");
177         }
178     }
179     return 0;
180 }