zoukankan      html  css  js  c++  java
  • {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)。
      1
      2
      3
      4
      5
      6
      7
      8
      9
     10
     11
     12
     13
     14
     15
     16
     17
     18
     19
     20
     21
     22
     23
     24
     25
     26
     27
     28
     29
     30
     31
     32
     33
     34
     35
     36
     37
     38
     39
     40
     41
     42
     43
     44
     45
     46
     47
     48
     49
     50
     51
     52
     53
     54
     55
     56
     57
     58
     59
     60
     61
     62
     63
     64
     65
     66
     67
     68
     69
     70
     71
     72
     73
     74
     75
     76
     77
     78
     79
     80
     81
     82
     83
     84
     85
     86
     87
     88
     89
     90
     91
     92
     93
     94
     95
     96
     97
     98
     99
    100
    101
    102
    103
    104
    105
    106
    107
    108
    109
    110
    111
    112
    113
    114
    115
    116
    117
    118
    119
    120
    121
    122
    123
    124
    125
    126
    127
    128
    129
    130
    131
    132
    133
    134
    #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;
    }
    
  • 相关阅读:
    js调用oc方法 UIWebView 跳 到另一个UIWebView
    可变cell,自适应cell,理解iOS 8中的Self Sizing Cells和Dynamic Type
    设置textField的placeholder字体的颜色,默认是灰色
    设置按钮(UIButton)为圆角矩形
    UITextView限制输入字数
    如何更好地限制一个UITextField的输入长度
    uitableview style类型为Grouped时,去掉上面空白的方法
    python note 01 计算机基础与变量
    Leetcode61.旋转链表
    Leetcode 430. 扁平化多级双向链表
  • 原文地址:https://www.cnblogs.com/jklongint/p/4749346.html
Copyright © 2011-2022 走看看