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Triathlon
Description Triathlon is an athletic contest consisting of three consecutive sections that should be completed as fast as possible as a whole. The first section is swimming, the second section is riding bicycle and the third one is running.
The speed of each contestant in all three sections is known. The judge can choose the length of each section arbitrarily provided that no section has zero length. As a result sometimes she could choose their lengths in such a way that some particular contestant would win the competition. Input The first line of the input file contains integer number N (1 <= N <= 100), denoting the number of contestants. Then N lines follow, each line contains three integers Vi, Ui and Wi (1 <= Vi, Ui, Wi <= 10000), separated by spaces, denoting the speed of ith contestant in each section.
Output For every contestant write to the output file one line, that contains word "Yes" if the judge could choose the lengths of the sections in such a way that this particular contestant would win (i.e. she is the only one who would come first), or word "No" if this is impossible.
Sample Input 9 10 2 6 10 7 3 5 6 7 3 2 7 6 2 6 3 5 7 8 4 6 10 4 2 1 8 7 Sample Output Yes Yes Yes No No No Yes No Yes Source |
这题坑了很久，总感觉有问题。
精度开到1e-18才过
1 /* *********************************************** 2 Author :kuangbin 3 Created Time :2013/8/18 19:47:45 4 File Name :F:2013ACM练习专题学习计算几何半平面交POJ1755_2.cpp 5 ************************************************ */ 6 7 #include <stdio.h> 8 #include <string.h> 9 #include <iostream> 10 #include <algorithm> 11 #include <vector> 12 #include <queue> 13 #include <set> 14 #include <map> 15 #include <string> 16 #include <math.h> 17 #include <stdlib.h> 18 #include <time.h> 19 using namespace std; 20 const double eps = 1e-18; 21 int sgn(double x) 22 { 23 if(fabs(x) < eps)return 0; 24 if(x < 0)return -1; 25 else return 1; 26 } 27 struct Point 28 { 29 double x,y; 30 Point(){} 31 Point(double _x,double _y) 32 { 33 x = _x; y = _y; 34 } 35 Point operator -(const Point &b)const 36 { 37 return Point(x - b.x, y - b.y); 38 } 39 double operator ^(const Point &b)const 40 { 41 return x*b.y - y*b.x; 42 } 43 double operator *(const Point &b)const 44 { 45 return x*b.x + y*b.y; 46 } 47 }; 48 //计算多边形面积 49 double CalcArea(Point p[],int n) 50 { 51 double res = 0; 52 for(int i = 0;i < n;i++) 53 res += (p[i]^p[(i+1)%n]); 54 return fabs(res/2); 55 } 56 //通过两点，确定直线方程 57 void Get_equation(Point p1,Point p2,double &a,double &b,double &c) 58 { 59 a = p2.y - p1.y; 60 b = p1.x - p2.x; 61 c = p2.x*p1.y - p1.x*p2.y; 62 } 63 //求交点 64 Point Intersection(Point p1,Point p2,double a,double b,double c) 65 { 66 double u = fabs(a*p1.x + b*p1.y + c); 67 double v = fabs(a*p2.x + b*p2.y + c); 68 Point t; 69 t.x = (p1.x*v + p2.x*u)/(u+v); 70 t.y = (p1.y*v + p2.y*u)/(u+v); 71 return t; 72 } 73 Point tp[110]; 74 void Cut(double a,double b,double c,Point p[],int &cnt) 75 { 76 int tmp = 0; 77 for(int i = 1;i <= cnt;i++) 78 { 79 //当前点在左侧，逆时针的点 80 if(a*p[i].x + b*p[i].y + c < eps)tp[++tmp] = p[i]; 81 else 82 { 83 if(a*p[i-1].x + b*p[i-1].y + c < -eps) 84 tp[++tmp] = Intersection(p[i-1],p[i],a,b,c); 85 if(a*p[i+1].x + b*p[i+1].y + c < -eps) 86 tp[++tmp] = Intersection(p[i],p[i+1],a,b,c); 87 } 88 } 89 for(int i = 1;i <= tmp;i++) 90 p[i] = tp[i]; 91 p[0] = p[tmp]; 92 p[tmp+1] = p[1]; 93 cnt = tmp; 94 } 95 double V[110],U[110],W[110]; 96 int n; 97 const double INF = 100000000000.0; 98 Point p[110]; 99 bool solve(int id) 100 { 101 p[1] = Point(0,0); 102 p[2] = Point(INF,0); 103 p[3] = Point(INF,INF); 104 p[4] = Point(0,INF); 105 p[0] = p[4]; 106 p[5] = p[1]; 107 int cnt = 4; 108 for(int i = 0;i < n;i++) 109 if(i != id) 110 { 111 double a = (V[i] - V[id])/(V[i]*V[id]); 112 double b = (U[i] - U[id])/(U[i]*U[id]); 113 double c = (W[i] - W[id])/(W[i]*W[id]); 114 if(sgn(a) == 0 && sgn(b) == 0) 115 { 116 if(sgn(c) >= 0)return false; 117 else continue; 118 } 119 Cut(a,b,c,p,cnt); 120 } 121 if(sgn(CalcArea(p,cnt)) == 0)return false; 122 else return true; 123 } 124 int main() 125 { 126 //freopen("in.txt","r",stdin); 127 //freopen("out.txt","w",stdout); 128 while(scanf("%d",&n) == 1) 129 { 130 for(int i = 0;i < n;i++) 131 scanf("%lf%lf%lf",&V[i],&U[i],&W[i]); 132 for(int i = 0;i < n;i++) 133 { 134 if(solve(i))printf("Yes "); 135 else printf("No "); 136 } 137 } 138 return 0; 139 }