pku 2187(最远点对)

 1 /*

 2 *  题目要求：求一组点中距离最远的一对的距离 

 3 *  解法：凸包+枚举 

 4 */

 5 

 6 #include <cmath>

 7 #include <cstdio>

 8 #include <cstdlib>

 9 #include <iostream>

10 

11 using namespace std;

12 

13 const int N = 50005;

14 

15 struct point {

16     int x;

17     int y;

18 }p[N], stack[N];

19 

20 int dis(point A, point B) {

21     return (A.x-B.x)*(A.x-B.x) + (A.y-B.y)*(A.y-B.y);

22 }

23 

24 int crossProd(point A, point B, point C) {

25     return (B.x-A.x)*(C.y-A.y) - (B.y-A.y)*(C.x-A.x);

26 }

27 

28 int cmp(const void *a, const void *b) {

29     point *c = (point *)a;

30     point *d = (point *)b;

31     int k = crossProd(p[0], *c, *d);

32     if (k<0 || (!k&&dis(p[0], *c)>dis(p[0], *d))) return 1;

33     return -1;

34 }

35 

36 int Graham(int n) {

37     int x = p[0].x;

38     int y = p[0].y;

39     int mi = 0;

40     for (int i=1; i<n; ++i) {

41         if (p[i].y<y || (p[i].y==y&&p[i].x<x)) {

42             x = p[i].x;

43             y = p[i].y;

44             mi = i;

45         }

46     } 

47     point tmp = p[mi];

48     p[mi] = p[0];

49     p[0] = tmp;

50     qsort(p+1, n-1, sizeof(point), cmp);

51     p[n] = p[0];

52     for (int i=0; i<2; ++i) stack[i] = p[i];

53     int top = 1;

54     for (int i=2; i<n; ++i) {

55         while (crossProd(stack[top-1], stack[top], p[i])<=0 && top>=1) --top;

56         stack[++top] = p[i];

57     }

58     return top;

59 }

60 

61 int solve(int n) {

62     int top = Graham(n);

63     int s, maxL = 0;

64     for (int i=0; i<top; ++i) {

65         for (int j=i+1; j<=top; ++j) {

66             s = dis(stack[i], stack[j]);

67             if (s > maxL) maxL = s;

68         }

69     }

70     return maxL;

71 }

72 

73 int  main() {

74     int n;

75     while (scanf("%d", &n) != EOF) {

76         for (int i=0; i<n; ++i) scanf ("%d%d", &p[i].x, &p[i].y);

77         int ans = solve(n);

78         printf ("%d\n", ans);

79     }

80     return 0;

81 }