【模板】最小正方形覆盖（坐标旋转加三分）

题目链接：https://vjudge.net/problem/POJ-3301

代码：

 1 /*************************************************************************
 2     > File Name: poj3301.cpp
 3 # File Name: poj3301.cpp
 4 # Author : xiaobuxie
 5 # QQ : 760427180
 6 # Email:760427180@qq.com
 7 # Created Time: 2019年10月23日 星期三 19时34分28秒
 8  ************************************************************************/
 9 
10 #include<iostream>
11 #include<cstdio>
12 #include<map>
13 #include<cmath>
14 #include<cstring>
15 #include<set>
16 #include<queue>
17 #include<vector>
18 #include<algorithm>
19 using namespace std;
20 typedef long long ll;
21 #define inf 0x3f3f3f3f
22 #define eps 1e-8
23 const double pi = acos(-1.0);
24 int sgn(double x){
25     if(fabs(x) < eps) return 0;
26     if(x<0) return -1;
27     return 1;
28 }
29 const int N = 49;
30 struct Point{
31     double x,y;
32     Point(){x=y=0;}
33     Point(double a,double b){x=a,y=b;}
34 }p[N];
35 int n;
36 double cal(double u){
37     double maxx,minx,maxy,miny;
38     maxx = maxy = -inf;
39     minx = miny = inf;
40     for(int i = 1;i<=n;++i){
41         double x = p[i].x*cos(u) - p[i].y*sin(u);
42         double y = p[i].x*sin(u) + p[i].y*cos(u);
43         maxx = max(maxx,x);
44         maxy = max(maxy,y);
45         minx = min(minx,x);
46         miny = min(miny,y);
47     }
48     return max( maxx - minx,maxy - miny);
49 }
50 int main(){
51     int T; scanf("%d",&T);
52     while(T--){
53         scanf("%d",&n);
54         for(int i = 1;i <= n; ++i) scanf("%lf %lf",&p[i].x,&p[i].y);
55         double l = 0,r = pi/2;
56         //三分旋转角度
57         while( r - l > eps){
58             double mid1 = (l+r)/2.0;
59             double mid2 = (mid1 + r)/2.0;
60             if( cal(mid1) > cal(mid2) ) l = mid1;
61             else r = mid2;
62         }
63         double ans = cal(l);
64         printf("%.2f\n",ans*ans);
65     }
66     return 0;
67 }

标签：double,ans,正方形,坐标,cal,include,mid2,模板,mid1
来源： https://www.cnblogs.com/xiaobuxie/p/11728385.html