UVA 10714 - Ants

最新推荐文章于 2020-10-29 22:01:57 发布

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本文介绍了一种基于贪心算法的解决方案，用于计算一群蚂蚁在一定长度的木杆上行走时，所有蚂蚁从木杆掉落所需的最早及最晚时间。通过分析蚂蚁之间的相对位置和可能的行走方向，算法确定了两种极端情况下的时间。该问题适用于竞赛编程和算法学习。

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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

贪心

最短时间:每只蚂蚁都往离pole近的点方向走,找出里面的最大值

最长时间:相邻两只蚂蚁的距离除2加上各自的离pole远的方向的距离,找出里面的最大值.

#include <iostream>
#include <memory.h>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

vector<int>antPos;
int len,n;
int main(){
	int t,i;

	scanf("%d",&t);
	while (t--)
	{
		antPos.clear();

		int minMaxTime=0,maxTime=0;

		scanf("%d%d",&len,&n);
		for (i=0;i<n;++i)
		{
			int temp;
			scanf("%d",&temp);
			minMaxTime=max(minMaxTime,min(temp,len-temp));//找出最小最大时间
			antPos.push_back(temp);
			maxTime=max(maxTime,max(antPos[i],len-antPos[i]));//找出离pole远的最大距离,不和蚂蚁相碰的情况
		}
		sort(antPos.begin(),antPos.end());
		for (int i=0;i<antPos.size()-1;++i)
		{
			int dis=(antPos[i+1]-antPos[i])/2;

			maxTime=max(maxTime,max(dis+(antPos[i]+dis),dis+(len-antPos[i]-dis)));//相碰的情况
		}
		printf("%d %d\n",minMaxTime,maxTime);
	}
	return 0;
}