POJ 1696 Space Ant 计算几何 叉积的应用

题目大意：平面内有一些点，我们要通过一些方式来走遍这所有的点，要求一个点只能走一次，只能向左转而不能向右转。求遍历这些点的顺序。

思路：数据范围是可以怎么搞都0ms的（n<=50,case<=100），所以只要有思路就可以了。

只能左转，想想好像有点像凸包啊。但是这个题要遍历所有的点，所以就把已经走过的点删掉，然后像凸包一样的往前走，每次找一个没走过的极角最小的点走，然后把它标记上。最后都走完就全部遍历完了。

CODE：

#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define MAX 110#define INF 0x7f7f7f7fusing namespace std;struct Point{int _id;double x,y;double alpha;Point(double _x,double _y):x(_x),y(_y) {}Point() {}Point operator -(const Point &a) {return Point(x - a.x,y - a.y);}bool operator <(const Point &a)const {return alpha < a.alpha;}void Read() {scanf("%d%lf%lf",&_id,&x,&y);}bool Cross(Point &a,Point &b) {Point p1 = *this - a,p2 = *this - b;return p1.x * p2.y - p2.x * p1.y < 0;}}point[MAX],now;int cases,points;int rubbish;inline void Initialize();int main(){for(cin >> cases;cases; --cases) {scanf("%d",&points);Initialize();for(int x,y,i = 1;i <= points; ++i) { point[i].Read();if(point[i].y < now.y)now = point[i];}printf("%d",points);now.x = 0;point[0] = now;for(int i = 1;i <= points; ++i) {for(int j = i + 1;j <= points; ++j)if(point[j].Judge(point[i],point[i - 1]))swap(point[i],point[j]);printf(" %d",point[i]._id);}puts("");}return 0;}inline void Initialize(){now = Point(0,INF);}