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 */