poj 1852 Ants（贪心）

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

题意：

有n只蚂蚁在长为l的木棒上爬行，当他们在木棒上相遇时，会朝相反的方向爬去，给你他们离起点的距离，当爬行到任意一个端点时，蚂蚁掉落，

求全部蚂蚁掉落的最长和最短时间

做法：

把全部蚂蚁看作一样的，当他们相遇时，看成相互穿过，这样直接计算就行

#include <cstdio>
#include <cstring>
#include <iostream>
#define INF 0x3f3f3f
using namespace std;

int main(){
    int i,t,Min,Max,k,n,l;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&l,&n);
        scanf("%d",&k);
        Min=min(k,l-k);
        Max=l-Min;
        for(i=2;i<=n;i++){
            scanf("%d",&k);
            Min=max(Min,min(k,l-k));
            Max=max(Max,max(k,l-k));
        }
        printf("%d %d\n",Min,Max);
    }
    return 0;
}