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。

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <map>
 #include <string>
 #include <vector>
 #include <set>
 #include <cmath>
 #include <ctime>
 #pragma comment(linker, "/STACK:102400000,102400000")
 #define lson (u << 1)
 #define rson (u << 1 | 1)
 #define rep(i, a, b) for(i = a; i < b; i++)
 #define reps(i, a, b, c) for(i = a; i < b; i += c)
 #define repi(i, a, b) for(i = a; i >= b; i--)
 #define cls(i, j) memset(i, j, sizeof i)
 using namespace std;
 typedef __int64 ll;
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const int maxn = 1e5 + ;
 const int maxm = ;
 const int inf = 0x3f3f3f3f;
 const ll linf = 0x3fffffffffffffff;
 const ll mod = 1e9 + ;

 double a[];
 int n;
 double maxi, mini;

 double dfs(int u, int pre, double limit, int fixed){
     //printf("%d %d\n",u, pre);
     if(u >= n - ) return limit;
     if(limit <= mini) return mini;
      double to_left = a[u] - a[pre];
      double to_right = a[u + ] - a[u];
      double tem;
      if(fixed){
         if(to_left >= limit) return dfs(u + , u, limit, );
         if(to_right >=  * limit || to_right == limit) return dfs(u + , u + , limit, );
         if(to_right > limit) return dfs(u + , u + , limit, );
         if(to_right < limit) return mini;
      }
      if(to_left >= limit) return dfs(u + , u, limit, );
      double tem1 = dfs(u + , u, to_left, );
      double tem2 = mini;
      if(to_right >=  * limit) tem2 = dfs(u + , u + , limit, );
      else{
         if(to_right <= limit) tem2 = dfs(u + , u + , to_right, );
         double tem3 = mini;
         if(limit >= to_right / ) tem3 = dfs(u + , u + , to_right / , );
         tem2 = max(tem2, tem3);
         tem3 = dfs(u + , u + , min(to_right, limit), );
         tem3 = max(tem2, tem3);
         if(limit < to_right / ) tem3 = dfs(u + , u + , limit, );
         tem2 = max(tem2, tem3);
      }
      return max(tem1, tem2);
 }

 int main(){
    // freopen("in.txt", "r", stdin);
     int T;
     scanf("%d", &T);
     while(T--){
         scanf("%d", &n);
         int i;
         rep(i, , n) scanf("%lf", &a[i]);
         sort(a, a + n);
         a[n] = ;
         mini = (double)linf;
         rep(i, , n) mini = min(mini, a[i] - a[i - ]);
         maxi = (double)linf;
         rep(i, , n - ) maxi = min(maxi, max(a[i + ] - a[i], a[i] - a[i - ]));
         double ans = dfs(, , maxi, );
         printf("%.3f\n", ans);
     }
     return ;
 }