【hdu1007】最近点对

http://acm.hdu.edu.cn/showproblem.php?pid=1007

分治法的经典应用，复杂度可以证明为nlognlogn

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <map>

 7 #include <stack>

 8 #include <string>

 9 #include <ctime>

10 #include <queue>

11 #define mem0(a) memset(a, 0, sizeof(a))

12 #define mem(a, b) memset(a, b, sizeof(a))

13 #define lson l, m, rt << 1

14 #define rson m + 1, r, rt << 1 | 1

15 #define eps 0.0000001

16 #define lowbit(x) ((x) & -(x))

17 #define memc(a, b) memcpy(a, b, sizeof(b))

18 #define x_x(a) ((a) * (a))

19 using namespace std;

20 #define LL __int64

21 #define DB double

22 #define pi 3.14159265359

23 #define MD 10000007

24 #define INF (int)1e9

25 struct Point {

26         double x, y;

27         Point(double x = 0, double y = 0):x(x), y(y) {}

28         void inp() {

29                 scanf("%lf%lf", &x, &y);

30         }

31         bool operator < (const Point &a) const {

32                 return x < a.x || x == a.x && y < a.y;

33         }

34 } pn[120000], tmp[120000];

35 int cmpy(Point i, Point j)

36 {

37         return i.y < j.y;

38 }

39 double dist(Point a, Point b)

40 {

41         return sqrt(x_x(a.x - b.x) + x_x(a.y - b.y));

42 }

43 double solve(int l, int r)

44 {

45         if(l == r) return INF;

46         int m = (l + r) >> 1;

47         double d1 = solve(l, m), d2 = solve(m + 1, r), d = min(d1, d2);

48         int nn = 0;

49         for(int i = l; i <= r; i++) {

50                 if(dist(pn[i], pn[m]) <= d) {

51                         tmp[++nn] = pn[i];

52                 }

53         }

54         sort(tmp + 1, tmp + nn, cmpy);

55         for(int i = 1; i < nn; i++) {

56                 double mind = INF;

57                 for(int j = i + 1; j <= nn && tmp[j].y - tmp[i].y <= d; j++) {

58                         mind = min(mind, dist(tmp[i], tmp[j]));

59                 }

60                 d = min(d, mind);

61         }

62         return d;

63 }

64 int main()

65 {

66         //freopen("input.txt", "r", stdin);

67         //freopen("output.txt", "w", stdout);

68         int n;

69         while(cin>> n, n) {

70                 for(int i = 1; i <= n; i++) pn[i].inp();

71                 sort(pn + 1, pn + 1 + n);

72                 double t = solve(1, n);

73                 printf("%.2f\n", t / 2);

74         }

75         return 0;

76 }

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