[POJ1852]Ants

 

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 12431		Accepted: 5462

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

Thinking

　　如果这个题使用穷竭搜索的话，时间复杂度O(2^n)；显然不行，我们可以采取一种更为巧妙的方法。如果两只蚂蚁碰头以后，他们互相折返，能不能把他们看成是向左的蚂蚁按照的向右的方向向右走，向右的亦是如此，实质上就是两个蚂蚁继续按照原方向行进。所以我们维护一个最小值，一个最大值即可。

 

var n,m,i,j,mint,maxt,t:longint;
    a:array[1..100000] of longint;
function min(x,y:longint):longint;
begin
    if x>y then exit(y) else exit(x);
end;
function max(x,y:longint):longint;
begin
    if x>y then exit(x) else exit(y);
end;
procedure main;
var i:longint;
begin
    readln(m,n);
    for i:=1 to n do
        read(a[i]);
    mint:=0;
    maxt:=0;
    for i:=1 to n do
        begin
            mint:=max(mint,min(a[i],m-a[i]));
            maxt:=max(maxt,max(a[i],m-a[i]));
        end;
    writeln(mint,' ',maxt);
end;
begin
    readln(t);
    for i:=1 to t do main;
end.