【思维】POJ 1852 Ants

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Ants

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 21211		Accepted: 8714

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

Source

Waterloo local 2004.09.19

题目大意是n只蚂蚁在长L的杆子上爬行，遇到端点就会掉落，相遇就会掉头，问你所有蚂蚁掉落的最长时间和最短时间。

这里不知道每只蚂蚁的初始爬行方向。

最短时间

每只蚂蚁都朝着最近的端点爬去的话，那么总体时间就是最短了。以中间为分界线，每只蚂蚁靠着哪边近就往哪边爬，对于所有的蚂蚁，那只最短时间最长的蚂蚁就是所有蚂蚁都掉落的最短时间。

最长时间

我们考虑一下两只蚂蚁相遇的情况，不难发现假如不考虑两只蚂蚁各自的差别的话，我们完全可以把相遇就掉头【考虑两只蚂蚁的不同】看成相遇就原样交错而过，那么全体的最长时间无疑就是所有蚂蚁中到自己最远的端点的时间了。

上AC代码

#include<stdio.h>

#define MAX_NUM 1000006

int a[MAX_NUM];
int Min,Max,L,n,T;

void solve(void)
{
    scanf("%d",&T);
    while(T--){
        scanf("%d",&L);
        scanf("%d",&n);
        for(int i = 0 ; i < n ; i++) scanf("%d",&a[i]);
        for(int i = 0 ; i < n ; i++){
            int temp = L - a[i] > a[i] ? a[i] : L - a[i] ;///距离两端最短的距离，就是这只蚂蚁掉下来最短的时间。
            Min = temp > Min ? temp : Min ;///比目前为止最短的时间还要长，更新
    
            temp = L - a[i] > a[i] ? L - a[i] : a[i];///找到该只蚂蚁距离两端最长的距离
            Max = temp > Max ? temp : Max;///比目前最长的时间还要长，更新
            
        }
        printf("%d %d\n",Min,Max);
        Min =0,Max = 0;///不要忘记初始化。
    }
}

int main(void)
{
    solve();

    return 0;
}