Graham凸包算法简介

凸包真是一个神奇的算法。。


概念

  • 凸包,我理解为凸多边形
  • 叉积 对于向量AB和向量BC,记向量AB*向量BC = AB * BC * sin ∠ABC,而叉积的绝对值其实就是S△ABC/2

对于平面上的一些点,我们要求凸包上所有的点,可以使用Graham算法 时间复杂度O(nlogn)


思路

先找到最左下的点,把其他的点按叉积排序。然后维护一个堆栈,每次利用叉积和栈顶比较判断当前枚举到的点是否是凸包上的点,是则弹出栈顶元素
具体算法Click here

  • 周长
    直接所有相邻两点距离相加

  • 面积
    多边形面积直接利用公式,用叉积计算


常熟巨大的丑陋代码

# include <stdio.h>
# include <stdlib.h>
# include <iostream>
# include <string.h>
# include <math.h>
# include <algorithm>
# define RG register
# define IL inline
# define ll long long
# define mem(a, b) memset(a, b, sizeof(a))
# define Min(a, b) (((a) > (b)) ? (b) : (a))
# define Max(a, b) (((a) < (b)) ? (b) : (a))
# define Sqr(a) ((a) * (a))
using namespace std;

const int MAXN = 50001;
int n, top;
struct Point{
    double x, y, len;
} p[MAXN], Point_A, s[MAXN]; //最左下的点 
//求叉积(向量ab,向量ac) 
IL double Cross(Point a, Point b, Point c){
    return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
}

IL double Dis(Point a, Point b){
    return sqrt(Sqr(a.x - b.x) + Sqr(a.y - b.y));
}
//极角排序 
IL bool Cmp(Point a, Point b){
    RG double x = Cross(Point_A, a, b);
    if(x > 0) return 1;
    else if(x < 0) return 0;
    a.len = Dis(Point_A, a);
    b.len = Dis(Point_A, b);
    return a.len < b.len;
}
//查找起始点,最左下
IL void Find(){
    RG int temp = 0;
    RG Point a = p[1];
    for(RG int i = 2; i <= n; i++)
        if(p[i].y < a.y || p[i].y == a.y && p[i].x < a.x){
            a = p[i];
            temp = i;
        }
    p[temp] = p[1];
    p[1] = a;
    Point_A = a;//保存起始点
}
//求凸包周长
IL double Length(){
    RG double sum = 0;
    for(RG int i = 1; i <= top; i++)
        sum += Dis(s[i - 1], s[i]);
    return sum;
}
//计算面积
IL double Area(){ 
    RG double sum = 0;
    for(RG int i = 1; i < top - 1; i++)
        sum += Cross(s[0], s[i], s[i + 1]);
    sum = fabs(sum) / 2;
    return sum;
}

IL void Graham(){
    Find();
    sort(p + 2, p + n + 1, Cmp);
    s[0] = p[1]; s[1] = p[2];
    for(RG int i = 3; i <= n; i++){
        while(Cross(s[top - 1], s[top], p[i]) <= 0 && top) top--;
        s[++top] = p[i];
    }
    s[++top] = p[1];
}

int main(){
    while(~scanf("%d", &n) && n){
        top = 1;
        for(RG int i = 1; i <= n; i++)
            scanf("%lf%lf", &p[i].x, &p[i].y);
        Graham();
        cout << top << " " << length() << " " << Area() << endl;
    }
    return 0;
}

板子题 1.Surround the Trees HDU - 1392 2.Cows POJ - 3348

原文地址:https://www.cnblogs.com/cjoieryl/p/8206404.html