格雷厄姆扫描法解凸壳问题

#include<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
using namespace std;

struct point
{
    double x;
    double y;
    double angle;
};
//判断点A，D是否在点B，C连线的两侧
bool different(point A,point B,point C,point D)
{
    double x1=A.x,x2=B.x,x3=C.x,x4=D.x;
    double y1=A.y,y2=B.y,y3=C.y,y4=D.y;
    if(x3==x2)
    {
        if((x1-x2)*(x4-x2)<0)
            return true;
        else 
            return false;
    }
    else
    {
        if((((x1-x2)*(y3-y2)+y2*(x3-x2))/(x3-x2))*(((x4-x2)*(y3-y2)+y2*(x3-x2))/(x3-x2))<0)
            return true;
        else return false;
    }    
}
double my_angle(point A,point B)
{
    if(A.x==B.x && A.y==B.y)
        return 0;
    return acos((B.x-A.x)/(sqrt((B.y-A.y)*(B.y-A.y)+(B.x-A.x)*(B.x-A.x))));
}
bool ismin(point A,point B)
{
    return A.y<B.y;
}
bool isAngle(point A,point B)
{
    return A.angle<B.angle;
}
int main()
{
    vector<point> num;
    int x,y;
    char c;
    while(cin>>x>>y)
    {
        point temp;
        temp.x=x;
        temp.y=y;
        temp.angle=0;
        num.push_back(temp);
        cin.get(c);
        if(c=='
')
            break;
    }
    sort(num.begin(),num.end(),ismin);
    point first=num[0];
    for(vector<point>::iterator iter=num.begin();iter!=num.end();iter++)
    {
        (*iter).angle=my_angle(first,*iter);
    }
    sort(num.begin(),num.end(),isAngle);
    vector<point> result;
    result.push_back(num[0]);
    result.push_back(num[1]);
    for(int i=2;i<num.size()-1;i++)
    {
        if(!different(num[i-2],num[i-1],num[i],num[i+1]))
            result.push_back(num[i]);
    }
    result.push_back(num[num.size()-1]);
    for(int p=0;p<result.size();p++)
    {
        cout<<result[p].x<<" "<<result[p].y<<endl;
    }
    system("pause");
    return 0;
}