题目链接
Quoit Design
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29547 Accepted Submission(s): 7741Problem DescriptionHave you ever played quoit in a playground? Quoit is a game in which flat rings are pitched at some toys, with all the toys encircled awarded.
In the field of Cyberground, the position of each toy is fixed, and the ring is carefully designed so it can only encircle one toy at a time. On the other hand, to make the game look more attractive, the ring is designed to have the largest radius. Given a configuration of the field, you are supposed to find the radius of such a ring.
Assume that all the toys are points on a plane. A point is encircled by the ring if the distance between the point and the center of the ring is strictly less than the radius of the ring. If two toys are placed at the same point, the radius of the ring is considered to be 0.InputThe input consists of several test cases. For each case, the first line contains an integer N (2 <= N <= 100,000), the total number of toys in the field. Then N lines follow, each contains a pair of (x, y) which are the coordinates of a toy. The input is terminated by N = 0.OutputFor each test case, print in one line the radius of the ring required by the Cyberground manager, accurate up to 2 decimal places.Sample Input2 0 0 1 1 2 1 1 1 1 3 -1.5 0 0 0 0 1.5 0Sample Output0.71 0.00 0.75
计算最小点对的思想是：分治
注意：在下面代码中，如果使用宏定义来定义来替换取小函数min，则ＴＬＥ。
Accepted Code:
1 /*************************************************************************
2 > File Name: 1007.cpp
3 > Author: Stomach_ache
4 > Mail: sudaweitong@gmail.com
5 > Created Time: 2014年06月10日 星期二 19时15分21秒
6 > Propose:
7 ************************************************************************/
8
9 #include <cmath>
10 #include <string>
11 #include <cstdio>
12 #include <vector>
13 #include <cstdlib>
14 #include <fstream>
15 #include <cstring>
16 #include <iostream>
17 #include <algorithm>
18 using namespace std;
19
20 #define INF (1000000000.0)
21 #define MAX_N (100010)
22 //#define X first
23 //#define Y second
24 //#define min(x, y) ((x) < (y) ? (x) : (y))
25 //typedef pair<double, double> pii;
26 struct pii {
27 double X, Y;
28 };
29
30 int n;
31 pii A[MAX_N], b[MAX_N];
32
33 double
34 min(double x, double y) {
35 return x < y ? x : y;
36 }
37
38 bool
39 compare_x(pii a, pii b) {
40 return a.X < b.X;
41 }
42
43 bool
44 compare_y(pii a, pii b) {
45 return a.Y < b.Y;
46 }
47
48 double
49 get_dist(pii a, pii b) {
50 return sqrt((a.X - b.X) * (a.X - b.X) + (a.Y - b.Y) * (a.Y - b.Y));
51 }
52
53 double
54 closest_pair(int low, int high) {
55 //if (high == low) return INF;
56 if (high - low == 1) return get_dist(A[low], A[high]);
57 if (high - low == 2) {
58 double d = get_dist(A[low], A[low+1]);
59 d = min(d, get_dist(A[low], A[high]));
60 d = min(d, get_dist(A[low+1], A[high]));
61 return d;
62 }
63 int m = (low + high) / 2;
64 double x = A[m].X;
65 double d = min(closest_pair(low, m), closest_pair(m+1, high)); // (1)
66
67 int len = 0;
68 for (int i = low; i <= high; i++) {
69 if (fabs(A[i].X - x) < d) {
70 b[len++] = A[i];
71 }
72 }
73 sort(b, b + len, compare_y);
74 for (int i = 0; i < len; i++) {
75 for (int j = i+1; j < len; j++) {
76 double dx = b[j].X - b[i].X;
77 double dy = b[j].Y - b[i].Y;
78 if (dy >= d) break;
79 d = min(d, sqrt(dx * dx + dy * dy));
80 }
81 }
82
83 return d;
84 }
85
86 void
87 solve() {
88 sort(A, A + n, compare_x);
89 printf("%.2f
", closest_pair(0, n - 1) * 0.5);
90 }
91
92 int
93 main(void) {
94 while (scanf("%d", &n) && n) {
95 for (int i = 0; i < n; i++)
96 scanf("%lf %lf", &A[i].X, &A[i].Y);
97
98 solve();
99 }
100
101 return 0;
102 }