poj 1852 Ants

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


题意:一群蚂蚁在长为Lcm的水平杆以1cm/s的速度行走，到杆的端点掉下，且不知道蚂蚁的走向。若两只蚂蚁相遇，则反向而行。求所有蚂蚁掉杆的最短时间跟最长时间。


思路：其实蚂蚁的走向不受相遇的影响，你可以把这两只蚂蚁看做互换真身了，哈哈，它们的运动方向没有改变了，就很容易了。


#include <iostream>
#include <cstdio>
using namespace std;
int t,p,len,n;
int main()
{
    scanf("%d",&t);
    while (t--)
    {
      int minnum = 0;
      int maxnum = 0;
      scanf("%d%d",&len, &n);
      for (int i = 0; i < n; i++)
      {
          scanf("%d",&p);
          int minn = min(p,len-p);
          int maxx = max(p,len-p);
          minnum = max(minn,minnum);
          maxnum = max(maxnum,maxx);
      }
      printf("%d %d\n",minnum,maxnum);
    }
    return 0;
}