• Gym 102460L Largest Quadrilateral


    Gym 102460L Largest Quadrilateral

    Largest Quadrilateral

    题意

    (n)个点从中选出四个点,使得面积最大

    Solution

    • 首先,肯定是求凸包,要求的点一定在凸包上。
    • 不难联想到凸包对每个边求最大三角形面积的问题,也就是旋转卡壳。
    • 可以将问题转化为,对凸包的每一个对角线(A_iA_j),求最大面积的两个三角形,( riangle{A_iA_jP}), ( riangle{A_iA_jQ}), 然后就可以枚举对角线,旋转卡壳算最大面积

    细节

    • 输出的格式
    • 凸包上应该留下共线的点
    #include <cstdio>
    #include <stack>
    #include <set>
    #include <cmath>
    #include <map>
    #include <time.h>
    #include <vector>
    #include <iostream>
    #include <string>
    #include <cstring>
    #include <algorithm>
    //#include <memory.h>
    #include <cstdlib>
    #include <queue>
    #include <iomanip>
    #include <cassert>
    // #include <unordered_map>
    #define P pair<int, int>
    #define LL long long
    #define LD long double
    #define PLL pair<LL, LL>
    #define mset(a, b) memset(a, b, sizeof(a))
    #define rep(i, a, b) for (int i = a; i < b; i++)
    #define PI acos(-1.0)
    #define random(x) rand() % x
    #define debug(x) cout << #x << " " << x << "
    "
    using namespace std;
    const int inf = 0x3f3f3f3f;
    const LL __64inf = 0x3f3f3f3f3f3f3f3f;
    #ifdef DEBUG
    const int MAX = 2e3 + 50;
    #else
    const int MAX = 1e6  + 50;
    #endif
    const int mod = 1e9+9;
    void file_read(){
    #ifdef DEBUG
        freopen("in", "r", stdin);
        // freopen("out", "w", stdout);
    #endif
    }
    
    template<typename type>
    struct Vec
    {
        type x, y;
        Vec() {}
        Vec(type x, type y) : x(x), y(y) {}
        friend istream & operator >> (istream &in, Vec &A) {
            in >> A.x >> A.y;
            return in;
        }    
        friend Vec operator - (const Vec &A, const Vec &B) {
            return Vec(A.x-B.x, A.y-B.y);
        }
        friend Vec operator + (const Vec &A, const Vec &B) {
            return Vec(A.x + B.x, A.y + B.y);
        }
        friend type det(const Vec &A, const Vec &B) {
            return A.x * B.y - A.y * B.x;
        }
        friend type dot(const Vec &A, const Vec &B) {
            return A.x * B.x + A.y * B.y;
        }
        friend bool operator < (const Vec &A, const Vec &B) {
            if(A.x != B.x) return A.x < B.x;
            return A.y < B.y;
        }  
        friend type area(const Vec &A, const Vec &B) {
            return abs(det(A, B));
        }
        friend type operator == (const Vec &A, const Vec &B) {
            return A.x == B.x and A.y == B.y;
        }
    };
    
    template<typename type>
    vector<Vec<type>> convex_hull(vector<Vec<type>> &pt) {
        sort(pt.begin(), pt.end());
        int n = pt.size();
        vector<Vec<type>> res(2*n);
        int k = 0 ;
        for(int i = 0; i < n; i++) {
            while(k > 1 and det(res[k-1]-res[k-2], pt[i]-res[k-1]) < 0) // <=会wa
                k--;
            res[k++] = pt[i];
        }
        for(int i = n-2, t = k; i >= 0; i--) {
            while(k > t and det(res[k-1]-res[k-2], pt[i]-res[k-1]) < 0) // <=会wa
                k--;
            res[k++] = pt[i];
        }
        res.resize(k-1);
        return res;
    }
    
    struct ModI
    {
        int i, n;
        ModI(int n ) : i(0), n(n) { assert(n > 0) ;} 
        ModI(int i, int n) : i(i%n), n(n) { assert(n > 0); }
        ModI operator ++ ( int ) {
            ModI row = ModI(i, n);
            i = (i + 1) % n;
            return row;
        }
        ModI operator + (int x) {
            ModI res = ModI(i, n);
            res.i = (res.i + x) % n;
            return res;
        }
        int operator = (int x) {
            return i = x;
        }
        bool operator < (int x) const {
            return i < x;
        }
        bool operator == (const ModI &other) const {
            return i  == other.i;
        }
        operator int () {
            return i;
        }
    };
    
    template<typename type>
    type area(const Vec<type> &A, const Vec<type> &B, const Vec<type> &C) {
        return area(A-B, A-C);
    }
    
    template<typename type>
    type rotateCalipers(vector<Vec<type>> pt) {
        int n = pt.size();
        type res = 0;
        for(int i = 0; i < pt.size(); i++) {
            ModI p1 = ModI(i+1, n);
            ModI p2 = ModI(i+3, n);
            for(ModI j = ModI(i+2, n); j+1 != i; j++) {
                while(p1+1 != j and area(pt[p1], pt[i], pt[j]) < area(pt[p1+1], pt[i], pt[j])) 
                    p1 ++;
                if(j == p2) p2++;
                while(p2+1 != i and area(pt[p2], pt[i], pt[j]) < area(pt[p2+1], pt[i], pt[j]))
                    p2 ++;
                auto cur = area(pt[p1], pt[i], pt[j]) + area(pt[p2], pt[i], pt[j]);
                res = max(res, cur);
            }
        }
        return res;
    }
    
    void out(LL ans) {
        if(ans & 1) {
            printf("%lld.5
    ", ans >> 1);
        }
        else {
            printf("%lld
    ", ans >> 1);
        }
    }
    
    int main() {
        file_read();
        int T;
        scanf("%d", &T);
        while (T--)
        {
            int n;
            scanf("%d", &n);
            vector<Vec<LL>> pt(n);
            for(int i = 0; i < n; i++) cin >> pt[i];
            auto ch = convex_hull(pt);
            if(ch.size() < 3) {
                printf("0
    ");
                continue;
            }
            if(ch.size() == 3) {
                LL ans = 0;
                LL A = area(ch[0], ch[1], ch[2]);
                for(auto p : pt) {
                    if(p == ch[0] or p == ch[1] or p == ch[2]) continue;
                    auto a = area(p, ch[1], ch[2]);
                    a = min(a, area(p, ch[0], ch[2]));
                    a = min(a, area(p, ch[0], ch[1]));
                    ans = max(ans, A-a);
                }
                out(ans);
                continue;
            }
            LL res = rotateCalipers(ch);
            out(res);
        }
        return 0;
    }
    
  • 相关阅读:
    安装SQL server 2016遇到问题
    Python:dictionary
    Python: tree data structure
    python3.4 data type
    Python 3.4 Library setup
    Python 3.4 send mail
    SDN实验---Ryu的应用开发(四)北向接口RESTAPI
    SDN实验---Ryu的应用开发(四)基于跳数的最短路径转发原理
    SDN实验---Ryu的应用开发(三)流量监控
    python---基础知识回顾(十)进程和线程(协程gevent:线程在I/O请求上的优化)
  • 原文地址:https://www.cnblogs.com/sidolarry/p/13568924.html
Copyright © 2020-2023  润新知