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  • {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=0;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?1:-1;
    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?1:-1;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 + 7;
    const double EPS = 1e-12;
    
    /* -------------------------------------------------------------------------------- */
    
    const int maxn = 1e2 + 7;
    
    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) / 2);
    }
    
    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 = 0; i < n; i ++) {
                scanf("%d%d", &p[i].x, &p[i].y);
            }
            line.clear();
            for (int i = 0; i < n; i ++) {
                for (int j = i + 1; j < n; j ++) {
                    line.pb(Line(p[i], p[j]));
                }
            }
            sort(all(line));
            ll area = - 1;
            for (int i = 1; i < line.size(); i ++) {
                for (int j = i - 1; j >= 0 && equal(line[i], line[j]); j --) {
                    umax(area, Area(line[i], line[j]));
                }
            }
            if (area < 0) puts("No Eyes");
            else cout << area << ".0000" << endl;
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/jklongint/p/4749346.html
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