poj 1852 ants

Ants

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 21090		Accepted: 8682

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

题解：题目大意，计算n只蚂蚁落下杆子的最短时间和最长时间，虽然两只蚂蚁相遇必须掉头，其实可以不考虑，就当两只蚂蚁互相穿过考虑，没啥影响的。。。求最短时间，则让每只蚂蚁都从离自己最近的杆子边下去，取最大值，求最大值则从最远的一边下去，取最大值。不要用cin读入，会超时。。。

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int main(){
	int t;
	cin>>t;
	while(t--){
		int m,n;
		scanf("%d %d",&m,&n);
		int minn=0,maxx=0;
		for(int i=0,x;i<n;i++){
			scanf("%d",&x);
			minn=max(minn,min(x,m-x));
			maxx=max(maxx,max(x,m-x));
		}
		cout<<minn<<" "<<maxx<<endl;
	}
	return 0;
}