2018 ICPC Asia Singapore Regional A. Largest Triangle (计算几何)

题目链接：Kattis - largesttriangle

Description

Given \(N\) points on a \(2\)-dimensional space, determine the area of the largest triangle that can be formed using \(3\) of those \(N\) points. If there is no triangle that can be formed, the answer is \(0\).

Input

The first line contains an integer \(N (3≤N≤5000)\) denoting the number of points. Each of the next \(N\) lines contains two integers \(x\) and \(y (0≤x,y≤4?10^7)\). There are no specific constraints on these \(N\) points, i.e. the points are not necessarily distinct, the points are not given in specific order, there may be \(3\) or more collinear points, etc.

Output

Print the answer in one line. Your answer should have an absolute error of at most \(10^{?5}\).

Sample Input

7
0 0
0 5
7 7
0 10
0 0
20 0
10 10

Sample Output

100.00000

Source

2018 ICPC Asia Singapore Regional

Solution

题意

给出 \(N\) 个点，选择其中 \(3\) 个点组成三角形，求最大面积的三角形的面积，如果不能组成三角形，输出 \(0\)。

题解

最大面积的三角形一定在凸包上，所以先求凸包。

接下来在凸包上枚举三个点。直接三重循环肯定超时。

可以枚举凸包上的两个点，另外一个点根据面积的单调性枚举。时间复杂度 \(O(N^2)\)。

还有 \(O(NlogN)\) 的做法，可以参考这篇论文。

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double db;
const db eps = 1e-10;
const db pi = acos(-1.0);
const ll maxn = 5010;

inline int dcmp(db x) {
    if(fabs(x) < eps) return 0;
    return x > 0? 1: -1;
}

class Point {
public:
    double x, y;
    Point(double x = 0, double y = 0) : x(x), y(y) {}

    void input() {
        scanf("%lf%lf", &x, &y);
    }

    bool operator<(const Point &a) const {
        return (!dcmp(x - a.x))? dcmp(y - a.y) < 0: x < a.x;
    }

    bool operator==(const Point &a) const {
        return dcmp(x - a.x) == 0 && dcmp(y - a.y) == 0;
    }

    db dis2(const Point a) {
        return pow(x - a.x, 2) + pow(y - a.y, 2);
    }

    db dis(const Point a) {
        return sqrt(dis2(a));
    }

    db dis2() {
        return x * x + y * y;
    }

    db dis() {
        return sqrt(dis2());
    }

    Point operator+(const Point a) {
        return Point(x + a.x, y + a.y);
    }

    Point operator-(const Point a) {
        return Point(x - a.x, y - a.y);
    }

    Point operator*(double p) {
        return Point(x * p, y * p);
    }

    Point operator/(double p) {
        return Point(x / p, y / p);
    }

    db dot(const Point a) {
        return x * a.x + y * a.y;
    }

    db cross(const Point a) {
        return x * a.y - y * a.x;
    }
};

db area(Point A, Point B, Point C) {
    return abs((A - B).cross(A - C));
}

typedef vector<Point> Polygon;
Polygon Andrew(vector<Point> p) {
    int n = p.size(), cnt = 0;
    Polygon ans(2 * n);
    sort(p.begin(), p.end());
    for (int i = 0; i < n; ++i) {
        while (cnt >= 2 && (ans[cnt - 1] - ans[cnt - 2]).cross(p[i] - ans[cnt - 2]) < eps) {
            --cnt;
        }
        ans[cnt++] = p[i];
    }
    int t = cnt + 1;
    for (int i = n - 1; i > 0; --i) {
        while (cnt >= t && (ans[cnt - 1] - ans[cnt - 2]).cross(p[i - 1] - ans[cnt - 2]) < eps) {
            --cnt;
        }
        ans[cnt++] = p[i - 1];
    }
    ans.resize(cnt - 1);
    return ans;
}

vector<Point> p;

int main() {
    int n;
    scanf("%d", &n);
    for(int i = 0; i < n; ++i) {
        Point tmp;
        tmp.input();
        p.push_back(tmp);
    }
    p = Andrew(p);
    n = p.size();
    db ans = 0.0;
    if(n < 3) {
        printf("%.5lf\n", ans);
        return 0;
    }
    for(int i = 0; i < n; ++i) {
        int k = i + 2;
        for(int j = i + 1; j < n; ++j) {
            while(k + 1 < n && area(p[i], p[j], p[k]) < area(p[i], p[j], p[k + 1])) {
                ++k;
            }
            ans = max(ans, area(p[i], p[j], p[k]));
        }
    }
    printf("%.5lf\n", ans * 0.5);
    return 0;
}

原文地址：https://www.cnblogs.com/wulitaotao/p/11626009.html