URAL 1963 Kite 四边形求对称轴数

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 10

using 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 0

0 9999

9999 9999

9999 0



0 0

1 2

2 2

2 1



ans:

8 

2



*/