UVA - 10714 - Ants

UVA - 10714

Ants

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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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

Piotr Rudnicki

Source

Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Problem Solving Paradigms :: Greedy ::  Non Classical, Usually Harder
Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) ::  Volume 4. Algorithm Design
Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Problem Solving Paradigms ::  Greedy - Standard
Root :: Prominent Problemsetters ::  Piotr Rudnicki

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思路：只需要遍历所有蚂蚁，把每个蚂蚁走出木杆的最长时间，最短时间分别求出来，取其中的最大值，就是题目所问的所有蚂蚁离开木杆的最短时间和最长时间

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#define INF 0x3fffffff
using namespace std;

int main() {
	int T;
	scanf("%d", &T);
	while(T--) {
		int len, n, MA = 0, MI = 0;
		scanf("%d %d", &len, &n);
		for(int i = 0; i < n; i++) {
			int t;
			scanf("%d", &t);
			int ma = len - t;
			int mi = len - ma;
			if(ma < mi) swap(ma, mi);
			MA = max(MA, ma);
			MI = max(MI, mi);
		}
		
		printf("%d %d\n", MI, MA);
	}
	return 0;
}