poj 1852 ants

Ants

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 10577		Accepted: 4684

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

只要明白一点，两个蚂蚁相遇以后往回走，对整体没有影响，相当于没有相遇。下面用箭头表示蚂蚁演示一下：

----> <---- 两个蚂蚁相遇，按照题目的意思，往回走，变为<---- ----> 。而假设不考虑相遇，情况是<---- ----> 。

与相遇是一样的。

源代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include <queue>

using namespace std;
#define maxn 1000005
#define inf 0x7ffffff
int arr[maxn];
int n,t,m;
int main()
{
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        int sm = 0;
        int lar = 0 ;
        for(int i = 0; i < m; i++){
            scanf("%d",&arr[i]);
           int tmp = min(arr[i],n - arr[i]);
           sm = max(sm,tmp);
           lar = max(lar,n - tmp);
        }
        printf("%d %d\n",sm,lar);
    }
    return 0;
}