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;
}