{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;
}
posted @ 2015-08-22 00:01  jklongint  阅读(179)  评论(0)  编辑  收藏  举报