2019牛客暑期多校训练营（第五场）I.three points 1(暴力几何)

题意：现在给你一个矩形边框 一个三角形的三边长 现在问你能否把三角形放入矩阵边框中 并且输出三个点的坐标

思路：我们可以发现如果一定有解 我们就可以让一个点在左下角(0，0)处 还有一个点在矩形边上 所以我们暴力排列一下三个点 找到一个合适的三角形然后输出

#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const int N = 3e5+7;
const int inf = 0x3f3f3f3f;
const double eps = 1e-6;
typedef long long ll;
const ll mod = 1e7+9;
int sgn(double x){
    if(fabs(x) < eps)return 0;
    if(x < 0)return -1;
    else return 1;
}
struct Point{
    double x,y;
    Point(){}
    Point(double _x,double _y){
        x = _x;
        y = _y;
    }
    void input(){
        scanf("%lf%lf",&x,&y);
    }
    void output(){
        printf("%.2f %.2f\n",x,y);
    }
    bool operator == (Point b)const{
        return sgn(x-b.x) == 0 && sgn(y-b.y) == 0;
    }
    bool operator < (Point b)const{
        return sgn(x-b.x)== 0?sgn(y-b.y)<0:x<b.x;
    }
    Point operator -(const Point &b)const{
        return Point(x-b.x,y-b.y);
    }
    //叉积
    double operator ^(const Point &b)const{
        return x*b.y - y*b.x;
    }
    //点积
    double operator *(const Point &b)const{
        return x*b.x + y*b.y;
    }
    //返回长度
    double len(){
        return hypot(x,y);//库函数
    }
    //返回长度的平方
    double len2(){
        return x*x + y*y;
    }
    //返回两点的距离
    double distance(Point p){
        return hypot(x-p.x,y-p.y);
    }
    Point operator +(const Point &b)const{
        return Point(x+b.x,y+b.y);
    }
    Point operator *(const double &k)const{
        return Point(x*k,y*k);
    }
    Point operator /(const double &k)const{
        return Point(x/k,y/k);
    }
    //`计算pa  和  pb 的夹角`
    //`就是求这个点看a,b 所成的夹角`
    //`测试 LightOJ1203`
    double rad(Point a,Point b){
        Point p = *this;
        return fabs(atan2( fabs((a-p)^(b-p)),(a-p)*(b-p) ));
    }
    //`化为长度为r的向量`
    Point trunc(double r){
        double l = len();
        if(!sgn(l))return *this;
        r /= l;
        return Point(x*r,y*r);
    }
    //`逆时针旋转90度`
    Point rotleft(){
        return Point(-y,x);
    }
    //`顺时针旋转90度`
    Point rotright(){
        return Point(y,-x);
    }
    //`绕着p点逆时针旋转angle`
    Point rotate(Point p,double angle){
        Point v = (*this) - p;
        double c = cos(angle), s = sin(angle);
        return Point(p.x + v.x*c - v.y*s,p.y + v.x*s + v.y*c);
    }
};
double w,h;
bool check(int x,int y,int z,double a,double b,double c){
    Point p[3];
    p[x]=Point(0.0,0.0);
    double angl;
    if(a<=w){
        p[y]=Point(a,0.0);
        angl=acos((a*a+b*b-c*c)/(2*a*b));
    }else{
        p[y]=Point(w,sqrt(a*a-w*w));
        angl=acos((a*a+b*b-c*c)/(2*a*b))+acos(w/a);
    }
    p[z]=Point(b*cos(angl),b*sin(angl));
    if(sgn(h-p[z].y)!=-1&&sgn(w-p[z].x)!=-1&&sgn(p[z].x)!=-1&&sgn(p[z].y)!=-1){
        printf("%.12f %.12f %.12f %.12f %.12f %.12f\n"
        ,p[0].x,p[0].y,p[1].x,p[1].y,p[2].x,p[2].y);
        return true;
    }
    return false;
}
int main(){
//    ios::sync_with_stdio(false);
//    cin.tie(0); cout.tie(0);
    int t; scanf("%d",&t);
    while(t--){
        scanf("%lf%lf",&w,&h);
        double a,b,c;
        scanf("%lf%lf%lf",&a,&b,&c);
        double g[5][5]={0};
        g[0][1]=a; g[0][2]=b; g[1][2]=c;
        g[1][0]=a; g[2][0]=b; g[2][1]=c;
        bool f=1;
        for(int i=0;i<3&&f;i++)
            for(int j=0;j<3&&f;j++){
                if(i==j) continue;
                for(int k=0;k<3&&f;k++){
                    if(k==i||k==j) continue;
                    if(check(i,j,k,g[i][j],g[i][k],g[j][k]))
                        f=0;
                }
            }
    }
}