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POJ 1852 Ants C语言版

POJ题解 Slyar 120浏览 0评论

文章作者:姜南(Slyar) 文章来源:Slyar Home (www.slyar.com) 转载请注明,谢谢合作。

挺脑筋急转弯的一道题...

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

Slyar:说下题目大意。n只蚂蚁在一根水平杆上,蚂蚁以1cm/s的速度沿着杆走,长度已知,每只蚂蚁可以向任意两个方向走。蚂蚁一旦走到末端就掉下去,两只蚂蚁一旦碰头就各自掉头向反方向走。问所有蚂蚁掉下去的最快时间和最慢时间。

我一开始也被"掉头"纠结了很长时间,后来画图的时候发现这个条件就是个鸡肋...蚂蚁没区别,速度也没区别,结果画图的时候发现掉头以后自己也不分不清2只蚂蚁...恍然大悟想到可以将蚂蚁看成根本没有掉头,反正速度一样...

这样,对每只蚂蚁求2个时间,较短的和较长的。然后对所有的时间分别找最大值。那么较短的最大值就是最快时间,较长的最大值就是最慢时间。

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  1. 《编程之美》上面有讲。。。
    Felix0218年前 (2009-06-04)回复