poj1673EXOCENTER OF A TRIANGLE

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据说这题是垂心。。数学太弱没有看出来，写了分朴实无华的代码。。

旋转三边得到图中的外顶点，然后连接三角形顶点求交点，交上WA。。觉得没什么错误就去看了下discuss,发现都在说精度问题，果断开始水，最后+了epsAC了。。

  1 #include <iostream>

  2 #include<cstdio>

  3 #include<cstring>

  4 #include<algorithm>

  5 #include<stdlib.h>

  6 #include<vector>

  7 #include<cmath>

  8 #include<queue>

  9 #include<set>

 10 using namespace std;

 11 #define N 100000

 12 #define LL long long

 13 #define INF 0xfffffff

 14 const double eps = 1e-8;

 15 const double pi = acos(-1.0);

 16 const double inf = ~0u>>2;

 17 struct Point

 18 {

 19     double x,y;

 20      Point(double x=0,double y=0):x(x),y(y) {}

 21 }p[5];

 22 typedef Point pointt;

 23 pointt operator + (Point a,Point b)

 24 {

 25     return Point(a.x+b.x,a.y+b.y);

 26 }

 27 pointt operator - (Point a,Point b)

 28 {

 29     return Point(a.x-b.x,a.y-b.y);

 30 }

 31 int dcmp(double x)

 32 {

 33     if(fabs(x)<eps) return 0;

 34     else return x<0?-1:1;

 35 }

 36 Point rotate(Point a,double rad)

 37 {

 38     return Point(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));

 39 }

 40 bool intersection1(Point p1, Point p2, Point p3, Point p4, Point& p)      // 直线相交

 41 {

 42     double a1, b1, c1, a2, b2, c2, d;

 43     a1 = p1.y - p2.y;

 44     b1 = p2.x - p1.x;

 45     c1 = p1.x*p2.y - p2.x*p1.y;

 46     a2 = p3.y - p4.y;

 47     b2 = p4.x - p3.x;

 48     c2 = p3.x*p4.y - p4.x*p3.y;

 49     d = a1*b2 - a2*b1;

 50     if (!dcmp(d))    return false;

 51     p.x = (-c1*b2 + c2*b1) / d;

 52     p.y = (-a1*c2 + a2*c1) / d;

 53     return true;

 54 }

 55 double cross(Point a,Point b)

 56 {

 57     return a.x*b.y-a.y*b.x;

 58 }

 59 double mul(Point p0,Point p1,Point p2)

 60 {

 61     return cross(p1-p0,p2-p0);

 62 }

 63 int main()

 64 {

 65     int n,i;

 66     cin>>n;

 67     while(n--)

 68     {

 69         for(i = 1; i <= 3 ; i++)

 70         scanf("%lf%lf",&p[i].x,&p[i].y);

 71         Point p1,p2,p3,p4;

 72         if(dcmp(mul(p[1],p[2],p[3]))>0)

 73         {

 74             p1 = rotate(p[3]-p[1],3*pi/2.0);

 75             p2 = rotate(p[2]-p[1],pi/2.0);

 76         }

 77         else

 78         {

 79             p1 = rotate(p[3]-p[1],pi/2.0);

 80             p2 = rotate(p[2]-p[1],3*pi/2.0);

 81         }

 82         p1.x+=p[1].x;

 83         p1.y+=p[1].y;

 84         p2.x+=p[1].x;

 85         p2.y+=p[1].y;

 86         p1.x = (p1.x+p2.x)/2;

 87         p1.y = (p1.y+p2.y)/2;

 88 

 89         if(dcmp(mul(p[2],p[1],p[3]))>0)

 90         {

 91             p3 = rotate(p[3]-p[2],3*pi/2.0);

 92             p4 = rotate(p[1]-p[2],pi/2.0);

 93         }

 94         else

 95         {

 96             p3 = rotate(p[3]-p[2],pi/2.0);

 97             p4 = rotate(p[1]-p[2],3*pi/2.0);

 98         }

 99         p3.x+=p[2].x;

100         p3.y+=p[2].y;

101         p4.x+=p[2].x;

102         p4.y+=p[2].y;

103         p3.x = (p3.x+p4.x)/2;

104         p3.y = (p3.y+p4.y)/2;

105         Point pp ;

106         intersection1(p1,p[1],p3,p[2],pp);

107         printf("%.4f %.4f\n",pp.x+eps,pp.y+eps);

108 

109     }

110     return 0;

111 }

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