题目链接：点击打开链接
题意：
问是否存在这样的四边形，若存在则输出四边形顶点
思路：
然后得到一个圆的2条弦，圆心就是4个顶点之一，剩下对称出来即可
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <set>
#include <stdlib.h>
using namespace std;
#define ST 1001
#define EN 1002
#define M 10000
#define inf 1000000
#define eps 1e-8
#define PR 1e-8
#define node Point
struct Point{//点是2维的
double x,y;
}p[50],aa[3],ans[4];
double Cross(Point p1,Point p2,Point p3,Point p4){//二维向量(p1p2)X(p3p4) 返回第三向量长度
double x1=p2.x-p1.x,y1=p2.y-p1.y;
double x2=p4.x-p3.x,y2=p4.y-p3.y;
return x1*y2-x2*y1; //为0表示 p1p2 与p3p4共线
//直线：不为0就是相交
}
bool On_Segment(Point p1,Point p2,Point p3){//p3点在 p1p2线段上
if(Cross(p1,p2,p1,p3)!=0)return false;
bool iny=(p1.y<=p3.y && p3.y<=p2.y)||(p1.y>=p3.y && p3.y>=p2.y);
bool inx=(p1.x<=p3.x && p3.x<=p2.x)||(p1.x>=p3.x && p3.x>=p2.x);
if(inx && iny)return true;
return false;
}
bool Segmentintersect(Point p1,Point p2,Point p3,Point p4){//p1p2 是否与 p3p4相交
double cross_1=Cross(p3,p4,p3,p1),cross_2=Cross(p3,p4,p3,p2);//cross_1 2必须一正一负且都不为0
double cross_3=Cross(p1,p2,p1,p3),cross_4=Cross(p1,p2,p1,p4);//cross_2 4必须一正一负且都不为0
//表示a线段 2点 在b线段 2侧
if(cross_1*cross_2<0 && cross_3*cross_4<0)return true;
//a线段端点在 b线段上 视情况取舍这种位置
// if(cross_1==0 && On_Segment(p3,p4,p1))return true;
// if(cross_2==0 && On_Segment(p3,p4,p2))return true;
// if(cross_3==0 && On_Segment(p1,p2,p3))return true;
// if(cross_4==0 && On_Segment(p1,p2,p4))return true;
return false;
}
#define N 105
#define eps 1e-8
double Abs(double x){return x>0?x:-x;}
bool equal(double a,double b){return Abs(a-b)<eps;}
node dui(node xx, node yy){
node now ;
now.x = 2*yy.x-xx.x;
now.y = 2*yy.y-xx.y;
return now;
}
double dis(node xx,node yy){
return sqrt((xx.x-yy.x)*(xx.x-yy.x)+(xx.y-yy.y)*(xx.y-yy.y));
}
node hehe(node a,node b,node c){
node now;
double a1=b.x-a.x, b1=b.y-a.y, c1=(a1*a1+b1*b1)/2;
double a2=c.x-a.x, b2=c.y-a.y, c2=(a2*a2+b2*b2)/2;
double d =a1*b2-a2*b1;
now.x = a.x+(c1*b2-c2*b1)/d;
now.y = a.y+(a1*c2-a2*c1)/d;
return now;
}
bool gongxian(node a,node b,node c){
if(!(equal((b.x+c.x)/2.0,a.x)&&equal((b.y+c.y)/2.0,a.y)))
swap(a,b);
if(!(equal((b.x+c.x)/2.0,a.x)&&equal((b.y+c.y)/2.0,a.y)))
return false;
return equal(dis(a,b),dis(a,c));
}
bool work(node a,node b,node c){
if(gongxian(a,b,c))return false;
node bb = dui(b,a);
if(gongxian(a,bb,c))return false;
ans[1] = hehe(a,bb,c);
ans[2] = dui(ans[1],a);
ans[3] = dui(ans[2],b);
ans[0] = dui(ans[1],c);
return Segmentintersect(ans[1],ans[3],ans[0],ans[2]);
}
int main(){
int T;scanf("%d",&T);
while(T--){
for(int i = 0; i < 3; i++)scanf("%lf %lf",&aa[i].x,&aa[i].y);
if(work(aa[0],aa[1],aa[2])||work(aa[1],aa[0],aa[2])||work(aa[2],aa[1],aa[0])){
puts("YES");
for(int i = 0; i < 4; i++){
printf("%.10lf %.10lf",ans[i].x,ans[i].y);
if(i==3)puts("");
else printf(" ");
}
}
else puts("NO"), puts("");
}
return 0;
}
CodeForces 23D Tetragon 给定凸四边形3条同边长的中点求4个顶点 计算几何,布布扣,bubuko.com
CodeForces 23D Tetragon 给定凸四边形3条同边长的中点求4个顶点 计算几何
原文地址：http://blog.csdn.net/qq574857122/article/details/36457879
