D. Tetragon

D. Tetragon

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You're given the centers of three equal sides of a strictly convex tetragon. Your task is to restore the initial tetragon.

Input

The first input line contains one number T — amount of tests (1 ≤ T ≤ 5·10⁴). Each of the followingT lines contains numbers x₁, y₁,x₂, y₂, x₃,y₃ — coordinates of different points that are the centers of three equal sides (non-negative integer numbers, not exceeding 10).

Output

For each test output two lines. If the required tetragon exists, output in the first lineYES, in the second line — four pairs of numbers — coordinates of the polygon's vertices in clockwise or counter-clockwise order. Don't forget, please, that the tetragon should be strictly convex, i.e. no 3 of its points lie on one line. Output numbers with 9 characters after a decimal point.

If the required tetragon doen't exist, output NO in the first line, and leave the second line empty.

Sample test(s)

Input

31 1 2 2 3 30 1 1 0 2 29 3 7 9 9 8

Output

NOYES3.5 1.5 0.5 2.5 -0.5 -0.5 2.5 0.5NO

解释：

设ABCD是所寻求的四边形，和K，L和M是相等的边AB，BC和CD，相应的中间点。让，M'是对称的点相对于L。然后BM'=CM = CL = BL =BK，我到M。 E。 B是三角形的KLM'外心。

了解B，我们可以得到整个四合院，使用的对称性点，K，L和M，然后检查它是否满足所有的条件。
需要注意的是，我们不知道哪一个给定的点是L，所以我们需要检查所有的3起案件。

代码：

#include <iostream>#include <sstream>#include <string>#include <vector>#include <deque>#include <queue>#include <set>#include <map>#include <algorithm>#include <functional>#include <utility>#include <cmath>#include <cstdlib>#include <ctime>#include <cstdio>using namespace std;#define REP(i,n) for((i)=0;(i)<(int)(n);(i)++)#define foreach(c,itr) for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)struct point{    double x,y;};#define eps 1.0E-9#define _abs(x) ((x)>0?(x):-(x))double area(point P, point Q, point R){    return ((Q.x - P.x) * (R.y - P.y) - (Q.y - P.y) * (R.x - P.x)) / 2.0;}point cross(point A, point B, point C, point D)  // AB - CD{    double a = -A.y + B.y, b = A.x - B.x, c = A.x * B.y - A.y * B.x, d = -C.y + D.y, e = C.x - D.x, f = C.x * D.y - C.y * D.x;    point ans = {(c*e - b*f) / (a*e - b*d), (c*d - a*f) / (b*d - a*e)};    return ans;}point func4(point &P, point &M){    point ans = {M.x * 2 - P.x, M.y * 2 - P.y};    return ans;}point func3(point &A, point &B){    double dx = B.x - A.x, dy = B.y - A.y;    point ans = {A.x + dx / 2.0 - dy / 2.0, A.y + dy / 2.0 + dx / 2.0};    return ans;}vector <point> func2(point A, point B, point C){    point D,E,F,G;    point H = func3(A,B), I = func3(B,A);    double dx = B.x - A.x, dy = B.y - A.y;    H.x += dx;    H.y += dy;    I.x += dx;    I.y += dy;    point J = func3(B,C), K = func3(C,B);    E = cross(H,I,J,K);    D = func4(E,C);    F = func4(E,B);    G = func4(F,A);    vector <point> ans;    double tmp;    bool pos = false, neg = false;    tmp = area(D,E,F);    if(tmp > -eps) pos = true;    if(tmp < eps) neg = true;    tmp = area(E,F,G);    if(tmp > -eps) pos = true;    if(tmp < eps) neg = true;    tmp = area(F,G,D);    if(tmp > -eps) pos = true;    if(tmp < eps) neg = true;    tmp = area(G,D,E);    if(tmp > -eps) pos = true;    if(tmp < eps) neg = true;    if(pos && neg) return ans;    ans.push_back(D);    ans.push_back(E);    ans.push_back(F);    ans.push_back(G);    return ans;}vector <point> func(point A, point B, point C){    vector <point> ans;    if(_abs(area(A,B,C)) < eps) return ans;    ans = func2(A,B,C);    if(!ans.empty()) return ans;    ans = func2(B,C,A);    if(!ans.empty()) return ans;    ans = func2(C,A,B);    if(!ans.empty()) return ans;    return ans;}int main(void){    int T,t,i;    point A,B,C;    scanf("%d",&T);    REP(t,T)    {        scanf("%lf%lf%lf%lf%lf%lf",&A.x,&A.y,&B.x,&B.y,&C.x,&C.y);        vector <point> ans = func(A,B,C);        if(!ans.empty()) printf("YES\n");        else printf("NO\n");        REP(i,ans.size())        {            printf("%.9f %.9f",ans[i].x,ans[i].y);            if(i != ans.size() - 1) printf(" ");        }        printf("\n");    }    return 0;}