URAL 1963 Kite 四边形求对称轴数

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题目链接： http://acm.timus.ru/problem.aspx?space=1&num=1963

题意，顺时针或逆时针给定4个坐标，问对称轴有几条，输出(对称轴数*2)

对于一条对角线，若他是对称轴，必须满足：另外2点关于这条对角线对称 =》 2条对角线互相垂直且与这条对角线形成的夹角相等

而对于每条边的中点，只需要证明中点相连后与其所在的边垂直即可，垂直且相等就可以证明是对称的

#include <iostream>#include <string>#include <cstring>#include <algorithm>#include <cstdio>#include <cctype>#include <queue>#include <stdlib.h>#include <cstdlib>#include <math.h>#include <set>#include <vector>#define eps 1e-8#define N 10using namespace std;struct point {double x,y;}p[N];bool eq(double a,double b){return (a-b>0?a-b:b-a)<eps;}bool chuizhi(point a,point b,point c,point d){if(!eq(a.x,b.x) && !eq(c.x,d.x))return eq(((a.y-b.y)*(c.y-d.y))/((a.x-b.x)*(c.x-d.x)),-1);if(eq(a.x,b.x))return eq(c.y,d.y);if(eq(c.x,d.x))return eq(a.y,b.y);}double dis(point a,point b){return sqrt((double)((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)));}double du(point a,point b,point c){double d1=dis(a,b),d2=dis(b,c),d3=dis(c,a);return (d1*d1+d3*d3-d2*d2)/d1*d3;}double k(point a,point b){//斜率if(a.x==b.x)return 456453454154.154674;return (a.y-b.y)/(a.x-b.x);}bool jiaodu(point a,point b,point c,point d)//a能否作为答案{//a,b对边 c,d对边point temp={(c.x+d.x)/2,(c.y+d.y)/2};if(eq(k(a,b),k(a,temp)))return true;return false;}void zhongdian(){p[4].x=(p[0].x+p[1].x)/2,p[4].y=(p[0].y+p[1].y)/2;p[5].x=(p[1].x+p[2].x)/2,p[5].y=(p[1].y+p[2].y)/2;p[6].x=(p[2].x+p[3].x)/2,p[6].y=(p[2].y+p[3].y)/2;p[7].x=(p[3].x+p[0].x)/2,p[7].y=(p[0].y+p[3].y)/2;}int main() {while (scanf ("%lf%lf", &p[0].x, &p[0].y) != EOF) {for (int i=1; i<4; ++i) scanf ("%lf%lf", &p[i].x, &p[i].y);int ans=0;zhongdian();if(chuizhi(p[0],p[2],p[1],p[3])&& jiaodu(p[0],p[2],p[1],p[3]))ans++;if(chuizhi(p[1],p[3],p[0],p[2])&& jiaodu(p[1],p[3],p[0],p[2]))ans++;if(chuizhi(p[4],p[6],p[0],p[1]) && chuizhi(p[4],p[6],p[2],p[3]))ans++;if(chuizhi(p[5],p[7],p[0],p[3]) && chuizhi(p[5],p[7],p[1],p[2]))ans++;printf("%d\n",ans*2);}return 0;}/*0 00 99999999 99999999 00 01 22 22 1ans:8 2*/