POJ 2007 Scrambled Polygon(极角排序)

题目链接：
POJ 2007 Scrambled Polygon

//向量a叉乘向量b小于0，说明向量b在向量a的右侧。
//极角排序
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
const int MAX_N=100;

int n=0;

struct Point {
    int x,y;
}point[MAX_N],origin;

bool cmp(Point a,Point b)
{
    int x1=a.x-origin.x,y1=a.y-origin.y;
    int x2=b.x-origin.x,y2=b.y-origin.y;
    return x2*y1-x1*y2 < 0; //计算b和原点的向量p1与a和原点的向量p2的叉积是否小于0，小于0说明p1在p2的右侧
}

int main()
{
    freopen("Bin.txt","r",stdin);
    scanf("%d%d",&origin.x,&origin.y);
    while(~scanf("%d%d",&point[n].x,&point[n].y)) {
        n++;
    }
    sort(point,point+n,cmp);
    printf("(%d,%d)\n",origin.x,origin.y);
    for(int i=0;iprintf("(%d,%d)\n",point[i].x,point[i].y);
    }
    return 0;
}