Space Ant ------极角排序

The most exciting space discovery occurred at the end of the 20th century. In 1999, scientists traced down an ant-like creature in the planet Y1999 and called it M11. It has only one eye on the left side of its head and just three feet all on the right side of its body and suffers from three walking limitations:
It can not turn right due to its special body structure.
It leaves a red path while walking.
It hates to pass over a previously red colored path, and never does that.

The pictures transmitted by the Discovery space ship depicts that plants in the Y1999 grow in special points on the planet. Analysis of several thousands of the pictures have resulted in discovering a magic coordinate system governing the grow points of the plants. In this coordinate system with x and y axes, no two plants share the same x or y.
An M11 needs to eat exactly one plant in each day to stay alive. When it eats one plant, it remains there for the rest of the day with no move. Next day, it looks for another plant to go there and eat it. If it can not reach any other plant it dies by the end of the day. Notice that it can reach a plant in any distance.
The problem is to find a path for an M11 to let it live longest.
Input is a set of (x, y) coordinates of plants. Suppose A with the coordinates (xA, yA) is the plant with the least y-coordinate. M11 starts from point (0,yA) heading towards plant A. Notice that the solution path should not cross itself and all of the turns should be counter-clockwise. Also note that the solution may visit more than two plants located on a same straight line.

Input
The first line of the input is M, the number of test cases to be solved (1 <= M <= 10). For each test case, the first line is N, the number of plants in that test case (1 <= N <= 50), followed by N lines for each plant data. Each plant data consists of three integers: the first number is the unique plant index (1..N), followed by two positive integers x and y representing the coordinates of the plant. Plants are sorted by the increasing order on their indices in the input file. Suppose that the values of coordinates are at most 100.
Output
Output should have one separate line for the solution of each test case. A solution is the number of plants on the solution path, followed by the indices of visiting plants in the path in the order of their visits.
Sample Input
2
10
1 4 5
2 9 8
3 5 9
4 1 7
5 3 2
6 6 3
7 10 10
8 8 1
9 2 4
10 7 6
14
1 6 11
2 11 9
3 8 7
4 12 8
5 9 20
6 3 2
7 1 6
8 2 13
9 15 1
10 14 17
11 13 19
12 5 18
13 7 3
14 10 16
Sample Output
10 8 7 3 4 9 5 6 2 1 10
14 9 10 11 5 12 8 7 6 13 4 14 1 3 2
题意：
1.找一条最长的逆时针方向路线（就是从最外面往里绕），要打印顺序。
分析：
2.输出的第一个数字是这一行的数字数量然后才是数字的顺序。
3.从Y轴坐标最小的点开始，如例1中是从8号（8，1）开始的。
4.极角排序时，以当前点为原点建立极坐标系，以极角的大小作为排序原则（这里十分不懂为什么用向量叉积来作为判断极角大小的依据，可能时数学没学好( ఠൠఠ )ﾉ）。不过在作叉积时是把第三维当作0的，如a(8,1)和b(2,4)是当作a(8,1,0)和b(2,4 0),所以Cross函数中才会只有x,y（可以自化简试试）。随着排序的进行未排序部分渐渐变少，数组中当前点之前的点都排好序了。
5.cmp()函数的参数要写成const类型参数，否则会报错的。
6.最开始的起点是y最小，最靠近原点（0，0）的点。

代码：

#include
#include
#include
#include
#include
#include
#include
#include 
#include 
#include
using namespace std;
typedef  long long ll;
typedef unsigned long long ull;
#define e 2.718281828459
#define INF 0x7fffffff
#pragma warning(disable:4996)
#define sf scanf
#define pf printf
const double pi = acos(-1.0);
//#define  eps 1e-9;


struct Point {
    int x, y;
    int dex;
    /*Point& operator=(Point a) {//编译器可按值传递方式自行实现，写不写都可以（因为没有指针，引用）
    x = a.x;
    y = a.y;
    dex = a.dex;
    }*/
};
Point temp;
int Cross(const Point& o, const Point& a, const Point& b) {//叉乘
    return (a.x - o.x)*(b.y - o.y) - (a.y - o.y)*(b.x - o.x);
}

double dis(const Point& o, const Point& a) {//点到原点距离
    return sqrt((double)((a.x - o.x)*(a.x - o.x) + (a.y - o.y)*(a.y - o.y)));

}
bool cmp(const Point& a, const Point& b) {
    int cho = Cross(temp, a, b);
    if (cho < 0)
        return false;
    if (cho > 0)
        return true;
    else {
        double dis1 = dis(temp, a);
        double dis2 = dis(temp, b);
        if (dis1 < dis2)
            return true;
        else
            return false;

    }


}

int main(void) {
    Point pnt[55];
    int M;
    int n;

    sf("%d", &M);
    while (M--) {
        sf("%d", &n);
        sf("%d %d %d", &pnt[0].dex, &pnt[0].x, &pnt[0].y);
        for (int i = 1; i < n; i++) {
            sf("%d %d %d", &pnt[i].dex, &pnt[i].x, &pnt[i].y);
            if (pnt[0].y > pnt[i].y) {
                temp = pnt[0];
                pnt[0] = pnt[i];
                pnt[i] = temp;
            }

        }
        pf("%d ", n);
        pf("%d ", pnt[0].dex);
        for (int i = 1; i < n; i++) {
            temp = pnt[i - 1];
            sort(pnt + i, pnt + n, cmp);
            pf("%d ", pnt[i].dex);

        }
        pf("\n");


    }

    return 0;
}