POJ1852 Ants

POJ1852 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 immediately 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

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2
3
4
5
2
10 3
2 6 7
214 7
11 12 7 13 176 23 191

Sample Output

1
2
4 8
38 207

Answer

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#include <cstdio>
#include <algorithm>

using namespace std;

int main() {
int iter;
scanf("%d", &iter);

while (iter--) {
int n, L;
scanf("%d %d", &L, &n);

int *x = new int[n];
for (int i = 0; i < n; ++i)
scanf("%d", &x[i]);

int minT = 0, maxT = 0;
for (int i = 0 ; i < n; ++i)
minT = max(minT, min(x[i], L - x[i]));

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

printf("%d %d\n", minT, maxT);

delete []x;
}
return 0;
}

Analysis

  1. 蚂蚁碰到后,各自掉头,其实因为速度不变,和没碰到没有任何区别,所以每只蚂蚁互不影响各自的状态,可以每只蚂蚁单独考虑。
  2. 在给定每只蚂蚁的坐标下,求最短时间,考虑一只蚂蚁,根据朝向有两种选择,哪个朝向到终点用时最短,则是该只蚂蚁的最短用时,如法炮制,求出所有蚂蚁各自的最短时长,最后取出最大值,是这群蚂蚁的最短用时。
  3. 最长时间同最短时间求法。
# POJ

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