Triangle（求凸包最大内接三角形）

题目描述

Given n distinct points on a plane, your task is to find the triangle that have the maximum area, whose vertices are from the given points.

输入

The input consists of several test cases. The first line of each test case contains an integer n, indicating the number of points on the plane. Each of the following n lines contains two integer xi and yi, indicating the ith points. The last line of the input is an integer −1, indicating the end of input, which should not be processed. You may assume that 1 <= n <= 50000 and −104 <= xi, yi <= 104 for all i = 1 . . . n.

输出

For each test case, print a line containing the maximum area, which contains two digits after the decimal point. You may assume that there is always an answer which is greater than zero.

样例输入 

3
3 4
2 6
2 7
5
2 6
3 9
2 0
8 0
6 5
-1

样例输出 

0.50
27.00

#include<bits/stdc++.h>//（求凸包最大内接三角形）O(n^2)

#define ll long long
#define N 50003
using namespace std;
const double pi = acos(-1.0);

struct node {
    double x, y;

    node(double X = 0, double Y = 0) {
        x = X, y = Y;
    }
} a[N], ch[N];

node operator-(node a, node b) {
    return node(a.x - b.x, a.y - b.y);
}

node operator+(node a, node b) {
    return node(a.x + b.x, a.y + b.y);
}

node operator*(node a, double t) {
    return node(a.x * t, a.y * t);
}

bool operator<(node a, node b) {
    return a.x < b.x || a.x == b.x && a.y < b.y;
}

int n, m;

double cross(node a, node b) {
    return a.x * b.y - a.y * b.x;
}

void convexhull() {
    sort(a + 1, a + n + 1);
    m = 0;
    if (n == 1) {
        ch[++m] = a[1];
        return;
    }
    for (int i = 1; i <= n; i++) {
        while (m > 1 && cross(ch[m - 1] - ch[m - 2], a[i] - ch[m - 2]) <= 0) m--;
        ch[m++] = a[i];
    }
    int k = m;
    for (int i = n - 1; i >= 1; i--) {
        while (m > k && cross(ch[m - 1] - ch[m - 2], a[i] - ch[m - 2]) <= 0) m--;
        ch[m++] = a[i];
    }
    m--;
}

double rotating() {
    if (m <= 2) return 0;
    if (m == 3) return fabs(cross(ch[1] - ch[0], ch[2] - ch[0])) / 2;
    double ans = 0;
    for (int i = 0; i < m; i++) {
        int j = (i + 1) % m;
        int k = (j + 1) % m;
        while (fabs(cross(ch[i] - ch[j], ch[i] - ch[k])) < fabs(cross(ch[i] - ch[j], ch[i] - ch[(k + 1) % m])))
            k = (k + 1) % m;
        while (i != j && k != i) {
            ans = max(ans, fabs(cross(ch[i] - ch[j], ch[i] - ch[k])));
            while (fabs(cross(ch[i] - ch[j], ch[i] - ch[k])) < fabs(cross(ch[i] - ch[j], ch[i] - ch[(k + 1) % m])))
                k = (k + 1) % m;
            j = (j + 1) % m;
        }
    }
    return ans / 2.0;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    while (cin >> n && n != -1) {
        for (int i = 1; i <= n; i++)cin >> a[i].x >> a[i].y;
        convexhull();
        double ans = rotating();
        cout << fixed << setprecision(2) << ans << endl;
    }
    return 0;
}