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

Source

Waterloo local 2004.09.19

 

 

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题意：

n只蚂蚁在长为L(cm)的杆上，速度为1cm/s，当蚂蚁爬至杆的顶端就会掉落，两只蚂蚁相遇时不能交错通过，只能朝着各自相反的方向爬回去。已知每只蚂蚁在杆上所处的位置，但不知道蚂蚁朝向，计算当所有的蚂蚁全部从杆落下需要的最短时间的最长时间。

 

思想：

 

先考虑最短时间：当所有的蚂蚁都朝着距离自己较近的端点爬所需时间最短，这种情况下不会发生相遇，也不可能在比此更短的时间内爬到端点。

 

再考虑最长时间：事实上，当两只蚂蚁相遇后，当它们保持原样交错通过也不会有什么问题。这样就可以把每只蚂蚁都当作独立的个体，要求最长的时间，实际上就是求蚂蚁距离杆的端点的最大值。

 

所以这题实际上就是在比大小 :)

  

Source Code:

#include <cstdio>

using namespace std;

int minVal(int a,int b){
    return a<b?a:b;
}

int maxVal(int a,int b){
    return a>b?a:b;
}

int Position[1000010];

int main()
{
    int n;
    int L,num;
    int minTime,maxTime;
    while(scanf("%d",&n)!=EOF){
        while(n--){
            scanf("%d%d",&L,&num);
            for(int i=0;i<num;++i)
                scanf("%d",&Position[i]);
            minTime=maxTime=0;
            for(int i=0;i<num;++i){
                minTime=maxVal(minTime,minVal(Position[i],L-Position[i]));
                maxTime=maxVal(maxTime,maxVal(Position[i],L-Position[i]));
            }
            printf("%d %d
",minTime,maxTime);
        }
    }
    return 0;
}