枚举+贪心 HDOJ 4932 Miaomiao's Geometry

题目传送门

 /*
     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
     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，
                 判断一下，取最大值。这题二分的内容少
 */
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 using namespace std;

 const int MAXN = ;
 const int INF = 0x3f3f3f3f;
 double x[MAXN];
 int n;

 bool check(double len)   {
     int last = -;
     for (int i=; i<=n; ++i)    {
         if (i ==  || i == n) continue;
         if (last == -) {
             if (x[i] - len >= x[i-]) {
                 last = -;  continue;
             }
             else if (x[i] + len <= x[i+])  {
                 last = ;   continue;
             }
             else    return false;
         }
         else if (last == ) {
             if (x[i-] + len == x[i])   {
                 last = -;   continue;
             }
             else if (x[i] - len >= x[i-] + len) {
                 last = -;  continue;
             }
             else if (x[i] + len <= x[i+])  {
                 last = ;   continue;
             }
             else    return false;
         }
     }
     return true;
 }

 int main(void)  {       //HDOJ 4932 Miaomiao's Geometry
     //freopen ("HDOJ_4932.in", "r", stdin);

     int T;  scanf ("%d", &T);
     while (T--) {
         scanf ("%d", &n);
         for (int i=; i<=n; ++i)    {
             scanf ("%lf", &x[i]);
         }
         sort (x+, x++n);

         double ans = 0.0;
         for (int i=; i<=n; ++i)    {
             if (check (x[i] - x[i-]))  ans = max (ans, x[i] - x[i-]);
             if (check ((x[i] - x[i-]) * 0.5))  ans = max (ans, (x[i] - x[i-]) * 0.5);
         }

         printf ("%.3f\n", ans);
     }

     return ;
 }