这题太坑姐了！！！！
给你正多边形的三个点，求包围它的最小矩形面积（矩形长宽必须平行于x或y）
按理是水题啊，求下三点的外接圆圆心（即正多边形的中心），然后随便找个点旋转向量即可。
有waterloo的测试数据，一直差了点精度，一直在看自己哪点损失精度了，可是旋转向量咋着也得用三角函数吧！！！
最后发现，我选的起始点是y最小的点，如果选输入的第一个点就和人家结果一摸一样了，坑姐啊！！！！神马破题！！
P.S. 测试数据http://plg1.cs.uwaterloo.ca/~acm00/010922/data/ D题
#include <queue> #include <stack> #include <math.h> #include <stdio.h> #include <stdlib.h> #include <iostream> #include <limits.h> #include <string.h> #include <string> #include <algorithm> using namespace std; struct point{double x,y;}; const double inf = 1e20; const double pi = acos(-1.0); point p[3]; const double eps = 1e-10; bool dy(double x,double y) { return x > y + eps;} // x > y bool xy(double x,double y) { return x < y - eps;} // x < y bool dyd(double x,double y) { return x > y - eps;} // x >= y bool xyd(double x,double y) { return x < y + eps;} // x <= y bool dd(double x,double y) { return fabs( x - y ) < eps;} // x == y point l2l_inst_p(point u1,point u2,point v1,point v2) { point ans = u1; double t = ((u1.x - v1.x)*(v1.y - v2.y) - (u1.y - v1.y)*(v1.x - v2.x))/ ((u1.x - u2.x)*(v1.y - v2.y) - (u1.y - u2.y)*(v1.x - v2.x)); ans.x += (u2.x - u1.x)*t; ans.y += (u2.y - u1.y)*t; return ans; } point circumcenter(point a,point b,point c) { point ua,ub,va,vb; ua.x = ( a.x + b.x )/2; ua.y = ( a.y + b.y )/2; ub.x = ua.x - a.y + b.y;//根据 垂直判断，两线段点积为0 ub.y = ua.y + a.x - b.x; va.x = ( a.x + c.x )/2; va.y = ( a.y + c.y )/2; vb.x = va.x - a.y + c.y; vb.y = va.y + a.x - c.x; return l2l_inst_p(ua,vb); } point Whirl(double cosl,double sinl,point a,point b) { b.x -= a.x; b.y -= a.y; point c; c.x = b.x * cosl - b.y * sinl + a.x; c.y = b.x * sinl + b.y * cosl + a.y; return c; } int main() { int n,ind = 1; double minx,miny,maxx,maxy; while( ~scanf("%d",&n) && n ) { minx = miny = inf; maxx = maxy = -inf; for(int i=0; i<3; i++) scanf("%lf%lf",&p[i].x,&p[i].y); point c = circumcenter(p[0],p[1],p[2]); point s = p[0],k; double a = 2*pi/n; for(int i=0; i<n; i++) { k = Whirl(cos(i*a),sin(i*a),c,s); if( xy(k.x,minx) ) minx = k.x; if( xy(k.y,miny) ) miny = k.y; if( dy(k.x,maxx) ) maxx = k.x; if( dy(k.y,maxy) ) maxy = k.y; } double ans = (maxx - minx)*(maxy - miny); printf("polygon %d: %.3lf\n",ind++,ans); } return 0; }