UVA - 10714 Ants（贪心）-优快云博客

Ants

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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 :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 4. Algorithm Design

Root :: Prominent Problemsetters :: Piotr Rudnicki

题目大意：
在一个lcm 竹竿上有n只蚂蚁，每只蚂蚁的能以1cm/s的速度移动，
现在题目告诉你竹竿的长度lcm和每只蚂蚁初始所在的位置。当两只蚂蚁头碰头相遇时，他们会以相反的方向运动。但是题目没告诉你每只蚂蚁初始运动的方向，现在要你求出所以蚂蚁都落下竹竿的最短的时间和最长的时间。

解析：贪心题，当两个蚂蚁相遇的情况可以不用考虑，因为可以想象为两只蚂蚁朝相同的方向进行接力。所以最长的时间就是所有蚂蚁中距离竹竿端点最长的距离，而最短的时间，就是所有蚂蚁中距离竹竿端点较短距离的最长距离（因为那只蚂蚁最后落下竹竿）

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 1000005;
int pos[N];
int len,n;
int main() {
	int t;
	scanf("%d",&t);
	while(t--) {
		scanf("%d %d",&len,&n);
		int maxa = -INF;
		int maxl;
		int mina = -INF;
		int minl;
		for(int i = 0; i < n; i++) {
			scanf("%d",&pos[i]);
			maxl = max(pos[i],len - pos[i]);
			maxa = max(maxa,maxl);
			minl = min(pos[i],len - pos[i]);
			mina = max(mina,minl);
		}
		printf("%d %d\n",mina,maxa);
	}
	return 0;
}