poj 1852 Ants 【思维】

最新推荐文章于 2021-01-19 15:45:20 发布

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博客介绍了POJ 1852 Ants的问题，包括问题描述、输入输出格式、示例及解题思路。关键在于理解当蚂蚁碰撞后各自回头时，可视为继续前进。最早时间取决于离杆子中点最近的蚂蚁，最晚时间是所有蚂蚁位置距离中点的最大差值。已提供通过验证的AC代码。

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Ants

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 12890		Accepted: 5635

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 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.

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

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

Sample Output

4 8
38 207

题意：在一个长度为L的水平杆上有N个爬行的蚂蚁，两只蚂蚁碰头后会各自向回走。现在给出蚂蚁的位置，蚂蚁的初始方向你可以任意选择，问你所有蚂蚁落在杆外的最早时间和最晚时间。

思路：把两只蚂蚁碰头后各自回头走->看做仍然向前走。最早时间是距离L/2最近的蚂蚁落在杆外的时间，最晚的时间是max(L-d, d)。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN (100000+1)
#define MAXM (100000)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
using namespace std;
int main()
{
    int t; Ri(t);
    W(t)
    {
        int N, L;
        Ri(L); Ri(N);
        int r = 0; int dis = INF, pos;
        for(int i = 1; i <= N; i++)
        {
            int d; Ri(d);
            r = max(r, max(L-d, d));
            if(abs(L/2-d) <= dis)
            {
                dis = abs(L/2-d);
                pos = d;
            }
        }
        int l = min(L-pos, pos);
        printf("%d %d\n", l, r);
    }
    return 0;
}