NUC1474 Ants【水题】-优快云博客

本文介绍了一个经典的编程问题——蚂蚁行走问题。该问题涉及一群蚂蚁在一根木棍上以恒定速度行走，当两只蚂蚁相遇时会转向反方向行走。文章提供了求解蚂蚁全部掉落木棍所需的最早与最晚时间的C++代码。

Ants

时间限制: 1000ms 内存限制: 65535KB

通过次数: 1总提交次数: 1

问题描述

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.

输入描述

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.

输出描述

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.

样例输入

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

样例输出

4 8
38 207

来源

Waterloo local 2004.09.19

问题分析：（略）

这个问题和《POJ1852 UVa10714 Ants【水题】》是同一个问题，代码直接用就AC了。

程序说明：参见参考链接。

参考链接：POJ1852 UVa10714 Ants【水题】

题记：程序做多了，不定哪天遇见似曾相识的。

AC的C++程序如下：

/* POJ1852 UVa10714 Ants */

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
    int t, l, n, pos, minans, maxans;

    scanf("%d", &t);;
    while(t--) {
        minans = 0;
        maxans = 0;

        scanf("%d%d", &l, &n);
        for(int i=1; i<=n; i++) {
            scanf("%d", &pos);

            minans = max(minans, min(pos, l - pos));

            maxans = max(maxans, max(pos, l - pos));
        }

        printf("%d %d\n", minans, maxans);
    }

    return 0;
}

转载于:https://www.cnblogs.com/tigerisland/p/7563668.html