先按题意找出最低点作为起始点，计算出起始向量。然后每次选择左转角度最小的点走。一定能走完 n 个点。
/**//*************************************************************************
Author: WHU_GCC
Created Time: 2007-8-13 19:38:56
File Name: pku1696.cpp
Description:
************************************************************************/
#include <iostream>
#include <cmath>
using namespace std;
#define out(x) (cout << #x << ": " << x << endl)
typedef long long int64;
const int maxint = 0x7FFFFFFF;
const int64 maxint64 = 0x7FFFFFFFFFFFFFFFLL;
template <class T> void show(T a, int n) 
{ for (int i = 0; i < n; ++i) cout << a[i] << ' '; cout << endl; }
template <class T> void show(T a, int r, int l) { for (int i = 0; i < r; ++i) show(a[i], l); cout << endl; }
typedef struct point_t
{
int num;
int x, y;
};
point_t operator -(const point_t &a, const point_t &b)
{
point_t ret;
ret.x = a.x - b.x;
ret.y = a.y - b.y;
return ret;
}
int64 det(const point_t &a, const point_t &b)
{
return int64(a.x) * b.y - int64(a.y) * b.x;
}
int64 dot(const point_t &a, const point_t &b)
{
return int64(a.x) * b.x + int64(a.y) * b.y;
}
double dist(const point_t &a)
{
return sqrt(double(a.x * a.x + a.y * a.y));
}
double angle(const point_t &a, const point_t &b)
{
double ret = acos(double(dot(a, b)) / (dist(a) * dist(b)));
return det(a, b) > 0 ? ret : -ret;
}
const int maxn = 100;
point_t p[maxn];
int used[maxn];
int main()
{
int ca;
int n;
for (scanf("%d", &ca); ca--;)
{
scanf("%d", &n);
for (int i = 0; i < n; i++)
scanf("%d%d%d", &p[i].num, &p[i].x, &p[i].y);
int miny = 0x7FFFFFFF, mini;
for (int i = 0; i < n; i++)
if (p[i].y < miny)
{
miny = p[i].y;
mini = i;
}
memset(used, 0, sizeof(used));
used[mini] = 1;
int now = mini;
point_t now_vector;
now_vector.x = 0;
now_vector.y = p[mini].y;
printf("%d", n);
for (int i = 1; i < n; i++)
{
printf(" %d", p[now].num);
double min_alpha = 1e200;
int minj;
for (int j = 0; j < n; j++) if (!used[j])
{
double t = angle(now_vector, p[j] - p[now]);
if (t < min_alpha)
{
min_alpha = t;
minj = j;
}
}
now_vector = p[minj] - p[now];
now = minj;
used[minj] = 1;
}
printf(" %d\n", p[now].num);
}
return 0;
}