{bzoj2338 [HNOI2011]数矩形 && NBUT 1453 LeBlanc}平面内找最大矩形

思路：

1. 枚举3个点，计算第4个点并判断是否存在，复杂度为O(N³logN)或O(N³α)
2. 考虑矩形的对角线，两条对角线可以构成一个矩形，它们的长度和中点必须完全一样，于是将所有线段按长度和中点排序，那么所有可能构成矩形的线段(对角线)一定在连续的区间内，顺序枚举即可，复杂度O(N²logN)。

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#pragma comment(linker, "/STACK:10240000") #include <map> #include <set> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include <vector> #include <cstdio> #include <string> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm>   using namespace std;   #define X first #define Y second #define pb push_back #define mp make_pair #define all(a) (a).begin(), (a).end() #define fillchar(a, x) memset(a, x, sizeof(a)) #define copy(a, b) memcpy(a, b, sizeof(a))   typedef long long ll; typedef pair<int, int> pii; typedef unsigned long long ull;   #ifndef ONLINE_JUDGE void RI(vector<int>&a,int n){a.resize(n);for(int i=;i<n;i++)scanf("%d",&a[i]);} void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?:-; while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> void print(const T t){cout<<t<<endl;}template<typename F,typename...R> void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> void print(T*p, T*q){int d=p<q?:-;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} #endif template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}   const double PI = acos(-1.0); const int INF = 1e9 + ; const double EPS = 1e-12;   /* -------------------------------------------------------------------------------- */   const int maxn = 1e2 + ;   struct Point { int x, y; Point(int x, int y) { this->x = x; this->y = y; } Point operator + (const Point &that) const { return Point(x + that.x, y + that.y); } Point operator - (const Point &that) const { return Point(x - that.x, y - that.y); } inline ll sqr(ll a) { return a * a; } ll dist(const Point &that) { return sqr(x - that.x) + sqr(y - that.y); } bool operator < (const Point &that) const { return x == that.x? y < that.y : x < that.x; } bool operator == (const Point &that) const { return x == that.x && y == that.y; } Point() {} };   struct Line { Point a, b, mid; ll len; Line(Point a, Point b) { this->a = a; this->b = b; mid = a + b; len = a.dist(b); } Line() {} bool operator < (const Line &that) const { return len == that.len? mid < that.mid : len < that.len; } }; vector<Line> line; Point p[maxn];   bool equal(const Line &a, const Line &b) { return a.len == b.len && a.mid == b.mid; }   ll cross(Point a, Point b) { return (ll)a.x * b.y - (ll)a.y * b.x; }   ll Area(const Line &a, Line &b) { return abs(cross(a.a - a.b, b.a - b.b) / ); }   int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGE int n; while (cin >> n) { for (int i = ; i < n; i ++) { scanf("%d%d", &p[i].x, &p[i].y); } line.clear(); for (int i = ; i < n; i ++) { for (int j = i + ; j < n; j ++) { line.pb(Line(p[i], p[j])); } } sort(all(line)); ll area = - ; for (int i = ; i < line.size(); i ++) { for (int j = i - ; j >= && equal(line[i], line[j]); j --) { umax(area, Area(line[i], line[j])); } } if (area < ) puts("No Eyes"); else cout << area << ".0000" << endl; } return ; }