题意:
t组样例，每组样例输入w， h， a， b， c.在坐标系中，0 <= x <= w, 0<=y<=h，
求出三个点X， Y， Z， 并且|XY| = a， |XZ| = b， |YZ| = c，求这三点坐标并依次输出
题解:
- 一个三角形在矩形中是合法的,那么就一定可以平移到矩阵的某个角,
- 然后让三角形的某个边和这个矩形的边重叠,
- 具体处理利用三角函数(不要用沟谷定理,减少误差)
#include <bits/stdc++.h> using namespace std; const double eps = 1e-8; double w, h; struct node{ double x, y; }no[3]; bool check(double a, double b, double c, int x, int y, int z){ no[x].x = 0.0; no[x].y = 0.0; if(a <= w){ no[y].x = a; no[y].y = 0.0; } else{ no[y].x = w; no[y].y = sqrt(a*a - w*w); } double ang = acos((a*a + b*b - c*c )*1.0/ (2*a*b)); ang = ang + atan(no[y].y*1.0 / no[y].x); no[z].x = b * cos(ang); no[z].y = b * sin(ang); if(no[z].x >= -eps && no[z].x <= w+eps && no[z].y >= -eps && no[z].y <= h+eps){ printf("%.12lf %.12lf %.12lf %.12lf %.12lf %.12lf\n", no[0].x, no[0].y, no[1].x, no[1].y, no[2].x, no[2].y); return true; } return false; } int main(){ int t; cin >> t; while(t--){ double a, b, c; scanf("%lf%lf%lf%lf%lf", &w, &h, &a, &b, &c); if(check(a, b, c, 0, 1, 2)) continue; if(check(a, c, b, 1, 0, 2)) continue; if(check(b, a, c, 0, 2, 1)) continue; if(check(b, c, a, 2, 0, 1)) continue; if(check(c, a, b, 1, 2, 0)) continue; if(check(c, b, a, 2, 1, 0)) continue; } return 0; }