POJ - 1852 Ants

                                                                                                                          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

 题解：

本题的难点，两只蚂蚁相遇后各自反向，相当于没有相遇，只是交换了身份，画图一目了然。蚂蚁A向左蚂蚁B向右相遇，蚂蚁A反向，此事只需把蚂蚁A看做是蚂蚁B。

<pre name="code" class="cpp">/* 
    coder: Shawn_Xue 
    date: 2015.3.16 
    result: AC 
    description: POJ - 1852 Ants
*/
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
#define MAXN 1000000+10

int ant[MAXN];

int main()
{

    int N;
    scanf("%d", &N);
    while (N--)
    {
        int l, n;
        scanf("%d%d", &l, &n);
        for (int i = 0; i < n; i++)
            scanf("%d", &ant[i]);
        sort(ant, ant+n);
        //  取距离两端最远距离为max
        int latest = max (l-ant[0], ant[n-1]); 
        double mid = 1.0 * l / 2;
        //  取距离中间最近距离为min
        double lmin = fabs(ant[0]-mid);
        int p = 0;
        for (int i = 1; i < n; i++)
            if (fabs(ant[i]-mid) < lmin)
            {
                lmin = fabs(ant[i]-mid);
                p = i;
            }
        int earliest = min(ant[p], l-ant[p]);
        printf("%d %d\n", earliest, latest);
    }
    return 0;
}