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 <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <climits>#include <cmath>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;i<n;i++){        printf("(%d,%d)\n",point[i].x,point[i].y);    }    return 0;}