POJ 1852 Ants

题意如下：

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

210 32 6 7214 711 12 7 13 176 23 191

Sample Output

4 838 207

        这道题乍一看情况很复杂，每只蚂蚁的朝向是不一定的，如果枚举每只蚂蚁的朝向，则为O(2^n)的复杂度，显然不合适，我们尝试打开一下思路。

        现在已知两只蚂蚁碰头之后会各自朝相反方向走，考虑到蚂蚁之间是没有个体差异的，如果我们将上述过程看做是两只蚂蚁碰头之后继续朝各自的方向行走，即忽略调转方向这一要求，事实上对结果也不会产生影响。所以我们只要枚举每只蚂蚁快速检查就可以了，复杂度O(n)。

#include <iostream>#include <cstdio>using namespace std;inline int read(void){int x=0,f=1; char ch=getchar();for(;ch<'0'||ch>'9';ch=getchar()) if(ch=='-') f=-1;for(;ch>='0'&&ch<='9';x=(x<<3)+(x<<1)+ch-'0',ch=getchar());return x*f;}const int maxn=1000009;int f[maxn];int main(int argc,char const *argv[]){int T=read();for(;T--;){int l=read(),n=read();for(int i=1;i<=n;i++) f[i]=read();int minn,maxx=minn=-0x3f3f3f;for(int i=1;i<=n;i++){minn=max(minn,min(f[i],l-f[i]));maxx=max(maxx,max(f[i],l-f[i]));}cout<<minn<<' '<<maxx<<endl;}return 0;}