Ants(POJ 1852)

Ants

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 24129		Accepted: 9652

Description

An army of ants walk on a horizontal pole of length l cm, each with a constant speed of 1 cm/s. When a walking ant reaches an end of the pole, it immediatelly falls off it. When two ants meet they turn back and start walking in opposite directions. We know the original positions of ants on the pole, unfortunately, we do not know the directions in which the ants are walking. Your task is to compute the earliest and the latest possible times needed for all ants to fall off the pole.

Input

The first line of input contains one integer giving the number of cases that follow. The data for each case start with two integer numbers: the length of the pole (in cm) and n, the number of ants residing on the pole. These two numbers are followed by n integers giving the position of each ant on the pole as the distance measured from the left end of the pole, in no particular order. All input integers are not bigger than 1000000 and they are separated by whitespace.

Output

For each case of input, output two numbers separated by a single space. The first number is the earliest possible time when all ants fall off the pole (if the directions of their walks are chosen appropriately) and the second number is the latest possible such time. 

Sample Input

2
10 3
2 6 7
214 7
11 12 7 13 176 23 191

Sample Output

4 8
38 207

• 题解：
  1. 最短时间很简单，所有蚂蚁往较近的一端走即可，这种情况下也不会发生两只蚂蚁相遇的情况，最短时间就是每只蚂蚁所走最短时间的最大值
  2. 对于最长时间，我们需要考虑两只蚂蚁相遇的情况，而事实上，两只蚂蚁相遇之后各自反向前进与两只蚂蚁相遇之后保持原样交错而过继续前进是等效的，因此可认为每只蚂蚁都是独立运动的，最长时间就是每只蚂蚁往较长一端走的时间的最大值
• 代码：

  #include <iostream>
  
  using namespace std;
  
  template <typename T>
  T minn(T, T);
  
  template int minn<int>(int, int);
  
  template <typename T>
  T maxn(T, T);
  
  template<> int maxn<int>(int, int);
  
  int main(int argc, char * argv[])
  {
      int k;
      for (cin >> k; k>0; k--)
      {
          int l, n;
          cin >> l >> n;
          int * a = new int[n];
          for (int i=0; i<n; i++) cin >> a[i];
          int tmin=0, tmax=0;
          for (int i=0; i<n; i++) 
          {
              tmin=maxn(tmin,minn(a[i],l-a[i]));
              tmax=maxn(tmax,maxn(a[i],l-a[i]));
          }
          cout << tmin << ' ' << tmax << endl;
      }
      return 0;
  }
  
  template <typename T>
  T minn(T x, T y)
  {
      if (x<y) return x;
      return y;
  }
  
  template <typename T>
  T maxn(T x, T y)
  {
      if (x>y) return x;
      return y;
  }
  
  template<> int maxn<int>(int x, int y)
  {
      if (x>y) return x;
      return y;
  }