poj 3301
题目：http://poj.org/problem?id=3301
题意：给出一些点的坐标，求一个最小的正方形来覆盖所有的点
用三分枚举转角即可，坐标转换公式： x’ = x * cosa - y * sina; y’ = y * cosa + x * sina;
View Code
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 using namespace std; 6 7 #define N 60 8 #define inf 100000000 9 #define eps 1e-8 10 double const pi=acos(-1.0); 11 struct node 12 { 13 double x,y; 14 }p[N],tp[N]; 15 int n; 16 node cal(node a,double key) 17 { 18 node temp; 19 temp.x = a.x * cos(key) - a.y * sin(key); 20 temp.y = a.y * cos(key) + a.x * sin(key); 21 return temp; 22 } 23 double ca(double key) 24 { 25 double minx,miny,maxx,maxy; 26 minx = miny = inf; 27 maxx = maxy = -inf; 28 for(int i = 0; i < n; i++) 29 { 30 tp[i] = cal(p[i],key); 31 minx = min(minx,tp[i].x); 32 miny = min(miny,tp[i].y); 33 maxx = max(maxx,tp[i].x); 34 maxy = max(maxy,tp[i].y); 35 } 36 double tx = maxx - minx; 37 double ty = maxy - miny; 38 return max(tx * tx,ty * ty); 39 } 40 int main() 41 { 42 int cas; 43 //freopen("data.txt","r",stdin); 44 scanf("%d",&cas); 45 while(cas--) 46 { 47 scanf("%d",&n); 48 for(int i=0; i<n; i++) 49 { 50 scanf("%lf%lf",&p[i].x,&p[i].y); 51 } 52 double l = 0.0; 53 double r = pi; 54 double lmid,rmid; 55 while(fabs(r - l) > eps) 56 { 57 lmid = (l * 2.0 + r) / 3.0; 58 rmid = (l + r * 2.0) / 3.0; 59 if(ca(lmid) > ca(rmid)) l = lmid; 60 else r = rmid; 61 } 62 printf("%.2lf\n",ca(l)); 63 } 64 return 0; 65 }
