1 /*
2 * 判断一堆点组成的形状是否为对称的
3 */
4
5 #include <cmath>
6 #include <cstdio>
7 #include <cstdlib>
8 #include <iostream>
9
10 using namespace std;
11
12 const int N = 10005;
13 const double EPS = 1e-8;
14
15 struct point {
16 double x, y;
17 }p[N], s;
18 int used[N];
19
20 int Rlcmp(double r1, double r2 = 0) {
21 if (fabs(r1 - r2) < EPS) return 0;
22 return r1 > r2 ? 1 : -1;
23 }
24
25 int cmp(const void *a, const void *b) {//按横坐标从小到大排序，横坐标相等时，按纵坐标从小到大排序
26 point *c = (point *)a;
27 point *d = (point *)b;
28 // if (c->x == d->x) return (int)(c->y - d->y);
29 // return (int)(c->x - d->x);
30 if (fabs(c->x - d->x) < EPS) return c->y > d->y ? 1 : -1;
31 return c->x > d->x ? 1 : -1;
32 }
33
34 bool solve(int n) {
35 int i;
36 s.x = s.y = 0;
37 for (i=0; i<n; ++i) {
38 used[i] = 1;
39 s.x += p[i].x, s.y += p[i].y;
40 }
41 s.x /= n, s.y /= n; //对称中心
42 for (i=0; i<n; ++i) { //将坐标系平移到点s
43 p[i].x -= s.x, p[i].y -= s.y;
44 if (Rlcmp(p[i].x) == 0 && Rlcmp(p[i].y) == 0) used[i] = 0;//判断跟点s重合的点，去掉
45 }
46 qsort(p, n, sizeof(point), cmp);
47 for (i=0; i<n; ++i) {//二分查找是否存在一个点，符合p-s = s-q
48 if (used[i]) {
49 used[i] = 0;
50 int ok = 0; //标志，初始为不存在
51 int left = 0;
52 int right = n - 1;
53 while (left <= right) {
54 int mid = (left + right) >> 1;
55 int u = Rlcmp(p[i].x + p[mid].x);
56 if (u == -1) left = mid + 1;
57 else if (u == 0) {
58 int v = Rlcmp(p[i].y + p[mid].y);
59 if (v == -1) left = mid + 1;
60 else if (v == 0) {
61 ok = 1; //存在
62 used[mid] = 0;
63 break;
64 }
65 else right = mid - 1;
66 }
67 else right = mid - 1;
68 }
69 if (ok == 0) break;//不存在
70 }
71 }
72 if (i == n) return true;
73 return false;
74 }
75
76 int main() {
77 int t;
78 scanf ("%d", &t);
79 while (t--) {
80 int n;
81 scanf ("%d", &n);
82 for (int i=0; i<n; ++i) scanf ("%lf%lf", &p[i].x, &p[i].y);
83 bool yes = solve(n);
84 if (yes) printf ("yes\n");
85 else printf ("no\n");
86 }
87 return 0;
88 }
