Saruman's Army(POJ 3069)

• 原题如下：
  Saruman's Army

  Time Limit: 1000MS	Memory Limit: 65536K
  Total Submissions: 14028	Accepted: 7038

  Description

  Saruman the White must lead his army along a straight path from Isengard to Helm’s Deep. To keep track of his forces, Saruman distributes seeing stones, known as palantirs, among the troops. Each palantir has a maximum effective range of R units, and must be carried by some troop in the army (i.e., palantirs are not allowed to “free float” in mid-air). Help Saruman take control of Middle Earth by determining the minimum number of palantirs needed for Saruman to ensure that each of his minions is within R units of some palantir.

  Input

  The input test file will contain multiple cases. Each test case begins with a single line containing an integer R, the maximum effective range of all palantirs (where 0 ≤ R ≤ 1000), and an integer n, the number of troops in Saruman’s army (where 1 ≤ n ≤ 1000). The next line contains n integers, indicating the positions x₁, …, x_n of each troop (where 0 ≤ x_i ≤ 1000). The end-of-file is marked by a test case with R = n = −1.

  Output

  For each test case, print a single integer indicating the minimum number of palantirs needed.

  Sample Input

  0 3
  10 20 20
  10 7
  70 30 1 7 15 20 50
  -1 -1

  Sample Output

  2
  4

  Hint

  In the first test case, Saruman may place a palantir at positions 10 and 20. Here, note that a single palantir with range 0 can cover both of the troops at position 20.

  In the second test case, Saruman can place palantirs at position 7 (covering troops at 1, 7, and 15), position 20 (covering positions 20 and 30), position 50, and position 70. Here, note that palantirs must be distributed among troops and are not allowed to “free float.” Thus, Saruman cannot place a palantir at position 60 to cover the troops at positions 50 and 70.

• 题解：我们可以从最左边的点开始考虑，对于这个点，其距离为R以内的区域里必须要有带有标记的点，带有标记的点一定在此点的右边，但究竟选它右边哪个点加上标记呢？这里我们的贪心策略是：从最左边的点开始，选择其距离为R以内的的区域里最远的点。因为更左的区域并没有覆盖的意义，所以应该覆盖更加靠右的点。加上第一个标记后，余下部分也用同样的办法处理，对于添加了标记的点的右侧相距超过R的下一个点，采用同样的办法找到其右侧相距R距离以内的最远的点添加标记即可，重复这一过程。
• 代码：

  #include <iostream>
  #include <algorithm>
  
  using namespace std;
  
  bool compare(int a, int b)
  {
      return a<b;
  }
  
  int n,r;
  int * x; 
  
  int main()
  {
      cin >> r >> n;
      while (r!=-1 && n!=-1)
      {
          int ans=0;
          x = new int[n];
          for (int i=0; i<n; i++) cin >> x[i];
          sort(x,x+n,compare);
          int i=0;
          while (i<n)
          {
              int s=x[i++];
              while(i<n && x[i]-s<=r) i++;
              int p=x[i-1];
              ans++;
              while(i<n && x[i]-p<=r) i++;
          }
          cout << ans << endl;
          delete[] x;
          cin >> r >> n;
      }
  }