# 题目

{% fold 点击显/隐题目 %}

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.

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 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.
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
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
{% endfold %}

# 代码

{% fold 点击显/隐代码 %}```cpp Space Ant https://github.com/OhYee/sourcecode/tree/master/ACM 代码备份
#include
#include
#include
#include
#include

using namespace std;
#define Log(format, ...) // printf(format, ##VA_ARGS)

/* 计算几何模板 */

const double eps = 1e-8;
const double INF = 9e50;
inline int sgn(const double &x) {
if (fabs(x) < eps)
return 0;
return x > 0 ? 1 : -1;
}

struct Vector {
double x, y;
int n;
Vector(double _x = 0, double _y = 0, int _n = 0) : x(_x), y(_y), n(_n) {}
bool operator==(const Vector &rhs) const {
return sgn(x - rhs.x) == 0 && sgn(y - rhs.y) == 0;
}
bool operator!=(const Vector &rhs) const { return !(*this == rhs); }
bool operator<(const Vector &rhs) const {
if (sgn(y - rhs.y) == 0)
return sgn(x - rhs.x) < 0;
return sgn(y - rhs.y) < 0;
}
Vector operator+(const Vector &rhs) const {
return Vector(x + rhs.x, y + rhs.y);
}
Vector operator-(const Vector &rhs) const {
return Vector(x - rhs.x, y - rhs.y);
}
double squre() const { return x * x + y * y; }
double distance() const { return sqrt(squre()); }
void print(bool flag = 0) const {
Log("(%.f %.f)", x, y);
if (flag)
Log("\n");
}
};
typedef Vector Point;
struct Segment {
Point a, b;
Segment() {}
Segment(Point _a, Point _b) : a(_a), b(_b) {}
Vector toVector() const { return b - a; }
double distance() const { return (b - a).distance(); }
void print(bool flag = 0) const {
a.print();
Log(" -> ");
b.print();
if (flag)
Log("\n");
}
};

/**
* 读入一个点的坐标
* @return 读入的点
*/
double x, y;
scanf("%lf%lf", &x, &y);
return Point(x, y);
}

/**
* 计算两个向量的叉积
* @param a 向量1
* @param b 向量2
* @return 叉积
*/
inline double Cross(const Vector &a, const Vector &b) {
return a.x * b.y - a.y * b.x;
}

/**
* 计算两个向量的点积
* @param a 向量1
* @param b 向量2
* @return 点积
*/
inline double Dot(const Vector &a, const Vector &b) {
return a.x * b.x + a.y * b.y;
}

/**
* 计算两点之间的距离
* @param a 线段L1
* @param b 线段L2
* @return 两点间的距离
*/
inline double Distance(const Point &a, const Point &b) {
return (a - b).distance();
}

/**
* 点和直线的关系
* @param p 目标点
* @param L 目标直线
* @return 1 在左侧,0 在直线上,-1在右侧
*/
inline int Point_Segment(const Vector &p, const Segment &L) {
return sgn(Cross(L.b - L.a, p - L.a));
}

/**
* 计算两个线段的位置关系
* @param L1 线段L1
* @param L2 线段L2
* @param p 返回交点坐标
* @return 2 重叠
1 相交
0 延长线相交
-1 平行
-2 共线不交
*/
inline int Segment_Segment(const Segment L1, const Segment L2,
Point *p = NULL) {
double a = L1.b.x - L1.a.x;
double b = L2.b.x - L2.a.x;
double c = L1.b.y - L1.a.y;
double d = L2.b.y - L2.a.y;
double f = a * d - b * c;
// 平行或重叠
if (sgn(f) == 0) {
if (Point_Segment(L1.a, L2)) {
// 平行
return -1;
} else {
// 共线
int len = max(max(Distance(L1.a, L2.a), Distance(L1.a, L2.b)),
max(Distance(L1.b, L2.a), Distance(L1.b, L2.b)));
if (sgn(len - L1.distance() - L2.distance()) > 0) {
// 共线不交
return -2;
} else {
// 重叠
return 2;
}
}
}
double g = L2.b.x - L1.a.x;
double h = L2.b.y - L1.a.y;
double t = (d * g - b * h) / f;
double s = (-c * g + a * h) / f;
if (p != NULL)
*p = Point(L1.a.x + t * a, L1.a.y + t * c);
// 在延长线上
if (t < 0 || t > 1 || s < 0 || s > 1)
return 0;
// 线段相交
return 1;
}

/**
* 判断点是否在多边形内部
* @param p 需要判断的点
* @param polygon 多边形点集,需要保证有序
* @param numberOfSide 多边形边数
* @return true 点在多边形内,false 点不在多边形内
*/
bool Point_Polygon(Point p, Point polygon[], int numberOfSide) {
bool ok =
Point_Segment(p, Segment(polygon[numberOfSide - 1], polygon[0])) >= 0;
for (int i = 1; i < numberOfSide && ok; ++i) {
if (!(Point_Segment(p, Segment(polygon[i - 1], polygon[i])) >= 0))
ok = false;
}
return ok;
}

/**
* 求点集的凸包
* @param p 点集,需要保证已经排序
* @param numOfPoint 点集内的点的个数
* @param vis 记录点对应的编号是否可以选择
* @param ans 返回的凸包
* @param begin 起始位置
* @return 返回凸包上点的个数
*/
int Convex_Hull(Point p[], int numOfPoint, bool vis[], Point ans[],
int begin = 0) {
int pos = begin - 1;
// 右链
for (int i = 0; i < numOfPoint; ++i) {
if (!vis[p[i].n]) {
while (pos > 0 &&
Point_Segment(p[i], Segment(ans[pos], ans[pos - 1])) > 0) {
vis[ans[pos--].n] = false;
}
ans[++pos] = p[i];
vis[ans[pos].n] = true;
}
}
// 左链
for (int i = numOfPoint - 2; i > 0; --i) {
if (!vis[p[i].n]) {
while (pos > 0 &&
Point_Segment(p[i], Segment(ans[pos], ans[pos - 1])) > 0) {
vis[ans[pos--].n] = false;
}
ans[++pos] = p[i];
vis[ans[pos].n] = true;
}
}
return pos + 1;
}

const int maxn = 105;

int n;
Point points[maxn];
Point ans[maxn];
bool vis[maxn];

void solve() {
int len = 0;
sort(points, points + n);
memset(vis, false, (n + 5) * sizeof(bool));

``````while (len != n) {
len = Convex_Hull(points, n, vis, ans, len);
Log("%d\n", len);
}

printf("%d", len);
for (int i = 0; i < len; ++i)
printf(" %d", ans[i].n);
printf("\n");
``````

}

int main() {
int T;
scanf("%d", &T);
while (T--) {
scanf("%d", &n);
for (int i = 0; i < n; ++i) {
int t;
scanf("%d", &t);
`{% endfold %}`