• POJ 1755 Triathlon (半平面交)


    Triathlon
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 4733   Accepted: 1166

    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 }
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  • 原文地址:https://www.cnblogs.com/kuangbin/p/3267057.html
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