Ants
Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu
Description
Problem B: Ants
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.
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.
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
Output for sample input
4 8 38 207
题意：一根棍子上有蚂蚁， 这些蚂蚁朝着一个方向以相同的速度（每秒一个单位长度）一直走（可能是往棍子的任意一端走），一旦蚂蚁碰到迎面走来的一直蚂蚁，那么这两只蚂蚁都会马上往相反方向走。求当所有蚂蚁都会走下棍子的，花费的最小的和最大时间。
思路：求最大时间即找出离两个端点最远的蚂蚁的最远位置，求最小时间求离两个端点最近的位置的蚂蚁的最远位置
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
using namespace std;
const int inf=0x3f3f3f3f;
int a[1000010];
int main(){
int T,n,m,i,j;
int maxx,minn;
scanf("%d",&T);
while(T--){
scanf("%d %d",&n,&m);
maxx=minn=-inf;
for(i=0;i<m;i++){
scanf("%d",&a[i]);
int left=a[i];
int right=n-a[i];
maxx = max(maxx, max(left,right));
minn = max(minn, min(left,right));
}
printf("%d %d\n", minn,maxx);
}
return 0;
}