pku 2526(Center of symmetry)

 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 }