POJ 1852(脑洞)

问题描述:

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

题目题意;题目给我们一根长L的木条，给了N只蚂蚁的位置，没有给定方向，如果蚂蚁相碰，那么就超反方向移动，蚂蚁每秒钟爬行1cm,问蚂蚁全部掉落木条的最小时间和蚂蚁最长能停留的时间。

题目分析;挑战程序设计竞赛里面讲到过这题，想法很巧妙。

如果蚂蚁A与蚂蚁B相碰，那么我们就把B蚂蚁现在看成A蚂蚁（方向与原来的A蚂蚁相同），A蚂蚁看成B蚂蚁，很巧妙的避开了相碰的难题，变成了蚂蚁单方向爬行。

代码如下:

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

int main()
{
    int t;
    scanf("%d",&t);
    while (t--) {
        int L,n,a;
        scanf("%d%d",&L,&n);
        int Min=0,Max=0;
        for (int i=0;i<n;i++) {
            scanf("%d",&a);
            int res=min(a,L-a);//求出每只蚂蚁的最快时间
            Min=max(Min,res);//然后求出最快时间的最大值
            Max=max(max(Max,a),max(Max,L-a));//直接求出最大值
        }
        printf("%d %d\n",Min,Max);
    }
    return 0;
}