POJ 2079 Triangle (平面点最大三角形凸包+旋转卡壳推荐)

最新推荐文章于 2019-10-31 21:12:05 发布

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#POJ #凸包 #旋转卡壳

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针对平面上给定的多个点，介绍如何通过计算几何的方法找到这些点中能构成的最大三角形面积。该算法首先利用凸包算法筛选出关键点集，然后通过枚举法找出最大的三角形。

Triangle

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 9424		Accepted: 2805

Description

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.

Input

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.

Output

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.

Sample Input

Sample Output

0.50
27.00

Source

Shanghai 2004 Preliminary

题目链接：http://poj.org/problem?id=2079

题目大意：平面上给n个点求任意三点能组成的最大三角形面积

题目分析；显然最大三角形的三个顶点都在凸包上(可以通过反证法)，求出凸包后枚举两个点，第三个点根据单调性直接逆时针旋转即可，时间复杂度O(n^2)，n方可做是因为坐标是整数点且范围到lim的纯净凸包上的点数量在sqrt(lim)附近，画个图就可以看出来，可以让横坐标每次x = x + 1，纵坐标y = y + x + 1，这样算下来，总的点数接近n*(n+1)/2 + n*(n+1)/2 + n = n^2 + 2n，n^2 + 2n < lim => n ~ sqrt(lim)，所以总点数也就在200多左右，n^2的旋转卡壳很快就能过。网上很多O(n)的代码都是错的，以下数据总有一组通不过(来自poj discuss)

答案：

n^2，282ms

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
int const MAX = 5e4 + 5;
int n;

struct POINT {
    int x, y;
}p[MAX], stk[MAX], base;

double getDist(POINT a, POINT b) {
    return sqrt(1.0 * (a.x - b.x) * (a.x - b.x) + 1.0 * (a.y - b.y) * (a.y - b.y));
}

double getCross(POINT p0, POINT p1, POINT p2) {
    return (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);
}

bool cmp(POINT p1, POINT p2) {
    if (getCross(base, p1, p2) == 0) {
        return getDist(base, p1) < getDist(base, p2);
    }
    if (getCross(base, p1, p2) > 0) {
        return true;
    }
    return false;
}

void getBase() {
    scanf("%d %d", &p[0].x, &p[0].y);
    base.x = p[0].x;
    base.y = p[0].y;
    int pos = 0;
    for (int i = 1; i < n; i ++) {
        scanf("%d %d", &p[i].x, &p[i].y);
        if (p[i].y < base.y || (p[i].y == base.y && p[i].x < base.x)) {
            base.x = p[i].x;
            base.y = p[i].y;
            pos = i;
        }
    }
    swap(p[0], p[pos]);
}

int main() {
    while (scanf("%d", &n) && n != -1) {
        getBase();
        sort(p + 1, p + n, cmp);
        stk[0] = p[0];
        stk[1] = p[1];
        int top = 1;
        for (int i = 2; i < n; i ++) {
            while (top > 0 && getCross(stk[top - 1], stk[top], p[i]) <= 0) {
                top --;
            }
            stk[++ top] = p[i];
        }
        double ans = 0;
        stk[++ top] = stk[0];
        for (int i = 0; i < top; i ++) {
            int k = 2;
            for (int j = i + 1; j <= top; j ++) {
                while(getCross(stk[i], stk[j], stk[k]) < getCross(stk[i], stk[j], stk[(k + 1) % top])) {
                    k = (k + 1) % top;
                }
                ans = max(ans, 0.5 * getCross(stk[i], stk[j], stk[k]));
            }
        }
        printf("%.2f\n", ans);
    }
}