【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

解题思路

【horizontal pole】 横杆

给出蚂蚁在横杆上的位置（不知道走向），求所有蚂蚁掉落的最短和最长时间。
题目说，蚂蚁相遇之后只能各自掉头走，但仔细想想，掉不掉头其实都一样。。蚂蚁之间又没差别(:з」∠)
最短时间就是让所有的蚂蚁朝离自己最近的一端走，最长时间就是朝最远的一端走。

参考代码

#include <cstdio>#include <algorithm>const int maxn = 1000010;using namespace std;int dis[maxn];int main(){    int T,i,l,n;    scanf("%d",&T);    while (T--){        scanf("%d%d",&l,&n);        for (i = 0;i < n;i++)            scanf("%d",&dis[i]);        int minT = 0,maxT = 0;;        for (i = 0;i < n;i++){            minT = max(minT,min(dis[i],l-dis[i]));//走离自己近的一端            maxT = max(maxT,max(dis[i],l-dis[i]));//走离自己远的一端        }//因为要保证所有的蚂蚁都掉落，所以都要取max        printf("%d %d\n",minT,maxT);            }    return 0;}