hdu4932 Miaomiao's Geometry

这是一道搜索题，我们很容易得到目标值的上下界，然后就只能枚举了。

就是将x轴上的点排序之后从左到右依次考察每个点，每个点要么在线段的左端点，要么在线段的右端点。

点编号从0到n-1，从编号为1的点开始，在枚举的过程中不断压缩上界，有一种情况需要特别讨论，即哪种一条线段恰好覆盖相邻两个点的。

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <map>
 5 #include <string>
 6 #include <vector>
 7 #include <set>
 8 #include <cmath>
 9 #include <ctime>
10 #pragma comment(linker, "/STACK:102400000,102400000")
11 #define lson (u << 1)
12 #define rson (u << 1 | 1)
13 #define rep(i, a, b) for(i = a; i < b; i++)
14 #define reps(i, a, b, c) for(i = a; i < b; i += c)
15 #define repi(i, a, b) for(i = a; i >= b; i--)
16 #define cls(i, j) memset(i, j, sizeof i)
17 using namespace std;
18 typedef __int64 ll;
19 const double eps = 1e-6;
20 const double pi = acos(-1.0);
21 const int maxn = 1e5 + 10;
22 const int maxm = 1050;
23 const int inf = 0x3f3f3f3f;
24 const ll linf = 0x3fffffffffffffff;
25 const ll mod = 1e9 + 7;
26
27 double a[55];
28 int n;
29 double maxi, mini;
30
31 double dfs(int u, int pre, double limit, int fixed){
32     //printf("%d %d\n",u, pre);
33     if(u >= n - 1) return limit;
34     if(limit <= mini) return mini;
35      double to_left = a[u] - a[pre];
36      double to_right = a[u + 1] - a[u];
37      double tem;
38      if(fixed){
39         if(to_left >= limit) return dfs(u + 1, u, limit, 1);
40         if(to_right >= 2 * limit || to_right == limit) return dfs(u + 2, u + 1, limit, 1);
41         if(to_right > limit) return dfs(u + 1, u + 1, limit, 1);
42         if(to_right < limit) return mini;
43      }
44      if(to_left >= limit) return dfs(u + 1, u, limit, 0);
45      double tem1 = dfs(u + 1, u, to_left, 0);
46      double tem2 = mini;
47      if(to_right >= 2 * limit) tem2 = dfs(u + 2, u + 1, limit, 0);
48      else{
49         if(to_right <= limit) tem2 = dfs(u + 2, u + 1, to_right, 1);
50         double tem3 = mini;
51         if(limit >= to_right / 2) tem3 = dfs(u + 2, u + 1, to_right / 2, 0);
52         tem2 = max(tem2, tem3);
53         tem3 = dfs(u + 1, u + 1, min(to_right, limit), 0);
54         tem3 = max(tem2, tem3);
55         if(limit < to_right / 2) tem3 = dfs(u + 2, u + 1, limit, 0);
56         tem2 = max(tem2, tem3);
57      }
58      return max(tem1, tem2);
59 }
60
61 int main(){
62    // freopen("in.txt", "r", stdin);
63     int T;
64     scanf("%d", &T);
65     while(T--){
66         scanf("%d", &n);
67         int i;
68         rep(i, 0, n) scanf("%lf", &a[i]);
69         sort(a, a + n);
70         a[n] = 0;
71         mini = (double)linf;
72         rep(i, 1, n) mini = min(mini, a[i] - a[i - 1]);
73         maxi = (double)linf;
74         rep(i, 1, n - 1) maxi = min(maxi, max(a[i + 1] - a[i], a[i] - a[i - 1]));
75         double ans = dfs(1, 0, maxi, 0);
76         printf("%.3f\n", ans);
77     }
78     return 0;
79 }

hdu4932 Miaomiao's Geometry