蚂蚁问题 - 爱码网

最近做了几个蚂蚁问题，还蛮有趣的。。。。。

蚂蚁问题第一弹：poj 1852 Ants：

Ants

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 12214		Accepted: 5366

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：一根木棒上有很多蚂蚁，给出这些蚂蚁初始时离木棒左端的距离，当然这些蚂蚁一开始是向左爬还是向右爬是未知的。由于木棒很细，因此如果两只蚂蚁碰到了它们会各自掉头继续爬，问这些蚂蚁全都掉下去最少和最多需要多少时间。

题解：首先可以确定两只蚂蚁相撞之后速度大小不变，只是换了一个方向而已，而且我们要求的是时间大小，具体是哪一只没差。那么可以把每一只蚂蚁相撞的情况看成是穿过。那么这一只蚂蚁要走的路的长短取决于方向即：L（木棒长）－X（初始位置）或者 X。

#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
    int T,L,n;
    cin>>T;
    while(T--)
    {
     int temp,cc,Min=-1e9,Max=-1e9;
     cin>>L>>n;
     for(int i=0;i<n;i++)
     {
         cin>>temp;
         cc=min(L-temp,temp);
         Min=max(cc,Min);
         cc=max(L-temp,temp);
         Max=max(cc,Max);
     }
     cout<<Min<<" "<<Max<<endl;
    }
    return 0;
}

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