二维凸包模板(Graham算法)

时间复杂度O(nlogn)

模板：

#include <cstdio>#include <cmath>#include <cstdlib>#include <algorithm>#define eps 1e-8#define MAXN 110using namespace std;struct Point{    double x, y;    Point(){}    Point(double X, double Y){        x = X; y = Y;    }};Point P[MAXN];int dcmp(double x){    if(fabs(x) < eps)        return 0;    else        return x < 0 ? -1 : 1;}Point operator - (Point A, Point B){    return Point(A.x-B.x, A.y-B.y);}Point operator + (Point A, Point B){    return Point(A.x+B.x, A.y+B.y);}Point operator * (Point A, double p){    return Point(A.x*p, A.y*p);;}double Cross(Point A, Point B){    return A.x*B.y - A.y*B.x;}double Dot(Point A, Point B){    return A.x*B.x + A.y*B.y;}double Dis(Point A, Point B){    return sqrt((A.x-B.x)*(A.x-B.x) + (A.y-B.y)*(A.y-B.y));}bool operator == (Point A, Point B){    return dcmp(A.x-B.x) == 0 && dcmp(A.y-B.y) == 0;}bool cmp(Point A, Point B)//按极角升序排序，若角度相等距离小的在前面{    double temp = Cross(A-P[0], B-P[0]);    if(dcmp(temp) > 0) return true;    if(dcmp(temp) == 0 && dcmp(Dis(P[0], A) - Dis(P[0], B)) < 0) return true;    return false;}int Stack[MAXN], top;//下标从0开始计数到topvoid input(int n){    scanf("%lf%lf", &P[0].x, &P[0].y);    double xx = P[0].x, yy = P[0].y;    int id = 0;//找到y坐标最小的点，若有多个选择x坐标最小的记录下标    for(int i = 1; i < n; i++)    {        scanf("%lf%lf", &P[i].x, &P[i].y);        if(P[i].y < yy || (P[i].y == yy && P[i].x < xx))        {            xx = P[i].x;            yy = P[i].y;            id = i;        }    }    Point T;    T = P[0]; P[0] = P[id]; P[id] = T;    sort(P+1, P+n, cmp);//极角排序}void Graham(int n){    if(n == 1){ top = 0; Stack[0] = 0;}    else if(n == 2)    {        top = 1;        Stack[0] = 0;        Stack[1] = 1;    }    else    {        for(int i = 0; i <= 1; i++)            Stack[i] = i;        top = 1;        for(int i = 2; i < n; i++)        {   //如果和上一条边成左旋关系，压栈继续；反之一直弹栈直到和栈顶两点的边成左转关系，压栈继续。            while(top > 0 && dcmp(Cross(P[Stack[top]]-P[Stack[top-1]], P[i]-P[Stack[top-1]])) <= 0) top--;            top++;            Stack[top] = i;        }    }}int main(){    int n;    input(n);    Graham(n);    return 0;}