题目传送门
1 /*
2 题意：有n个点，用相同的线段去覆盖，当点在线段的端点才行，还有线段之间不相交
3 枚举+贪心：有坑点是两个点在同时一条线段的两个端点上，枚举两点之间的距离或者距离一半，尽量往左边放，否则往右边放，
4 判断一下，取最大值。这题二分的内容少
5 */
6 #include <cstdio>
7 #include <algorithm>
8 #include <cstring>
9 #include <cmath>
10 using namespace std;
11
12 const int MAXN = 55;
13 const int INF = 0x3f3f3f3f;
14 double x[MAXN];
15 int n;
16
17 bool check(double len) {
18 int last = -1;
19 for (int i=1; i<=n; ++i) {
20 if (i == 1 || i == n) continue;
21 if (last == -1) {
22 if (x[i] - len >= x[i-1]) {
23 last = -1; continue;
24 }
25 else if (x[i] + len <= x[i+1]) {
26 last = 1; continue;
27 }
28 else return false;
29 }
30 else if (last == 1) {
31 if (x[i-1] + len == x[i]) {
32 last = -1; continue;
33 }
34 else if (x[i] - len >= x[i-1] + len) {
35 last = -1; continue;
36 }
37 else if (x[i] + len <= x[i+1]) {
38 last = 1; continue;
39 }
40 else return false;
41 }
42 }
43 return true;
44 }
45
46 int main(void) { //HDOJ 4932 Miaomiao's Geometry
47 //freopen ("HDOJ_4932.in", "r", stdin);
48
49 int T; scanf ("%d", &T);
50 while (T--) {
51 scanf ("%d", &n);
52 for (int i=1; i<=n; ++i) {
53 scanf ("%lf", &x[i]);
54 }
55 sort (x+1, x+1+n);
56
57 double ans = 0.0;
58 for (int i=2; i<=n; ++i) {
59 if (check (x[i] - x[i-1])) ans = max (ans, x[i] - x[i-1]);
60 if (check ((x[i] - x[i-1]) * 0.5)) ans = max (ans, (x[i] - x[i-1]) * 0.5);
61 }
62
63 printf ("%.3f
", ans);
64 }
65
66 return 0;
67 }