POJ 1852 Ants （思维技巧 + 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

Sample Output

4 8
38 207

Source

Waterloo local 2004.09.19

题意：n只蚂蚁以1cm/s的速度在长为L的棍子上爬，给出n只蚂蚁的初始距离左端的距离x，但不知道他们的爬行方向。规定两只蚂蚁相遇时，不能交错通过，只能个自反向爬回去。问所有蚂蚁落下棍子的最短时间和最长时间。

解析：这里有一个小技巧：当两只蚂蚁相遇时，我们可以认为两只蚂蚁可以直接通过，互不影响，这样就相当于两只蚂蚁跟原来的真正身份互换了一下，但是它们又没有明确要求，所以这是不影响结果的。然后再利用贪心的思想，按照对我们所求结果最有利的方向去安排个只蚂蚁的爬行方向即可。

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <stack>
using namespace std;
#define INF 0x7fffffff
#define LL long long
#define MID(a, b)  a+(b-a)/2
const int maxn = 1000000 + 10;

int main(){
    #ifdef sxk
        freopen("in.txt", "r", stdin);
    #endif // sxk

    int L, n, t;
    int a[maxn];
    scanf("%d", &t);
    while(t--){
        scanf("%d%d", &L, &n);
        for(int i=0; i<n; i++) scanf("%d", &a[i]);

        int mn = 0;
        for(int i=0; i<n; i++)
            mn = max(mn, min(a[i], L - a[i]));

        int mx = 0;
        for(int i=0; i<n; i++)
            mx = max(mx, max(a[i], L - a[i]));

        printf("%d %d\n", mn, mx);
    }
    return 0;
}