【弹性碰撞问题】POJ 1852 Ants

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

有n只蚂蚁在木棍上爬行，每只蚂蚁的速度都是每秒1单位长度，现在给你所有蚂蚁初始的位置（蚂蚁运动方向未定），蚂蚁相遇会掉头反向运动，让你求出所有蚂蚁都·掉下木棍的最短时间和最长时间。

分析

看到这个题，一开始的想法应该都是直接暴搜氵分，但是注意一下数据范围，肯定是过不了的，所以要换一种思想。 因为是同时出发的，相遇时的两只蚂蚁用的时间是相同的，我们可以无视蚂蚁的区别，当两只蚂蚁相遇时它们保持原样交错而行。这样每只蚂蚁都是独立运动的，那么只要找每只蚂蚁掉下去的时间就行了。

代码

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 1e6+10;
int a[maxn],ansx,ansd,L,n;
void ans1(){
    int Min;
    ansx = -1;
    for(int i=0;i<n;++i){
        Min = min(a[i],L-a[i]);
        if(Min>ansx)
            ansx = Min;
    }
    printf("%d ",ansx);
}
void ans2()
{
    int Max;
    ansd=-1;
    for(int i=0;i<n;++i)
    {
        Max=max(a[i],L-a[i]);
        if(Max>ansd)
            ansd=Max;
    }
    printf("%d
",ansd);
}

int main(){

    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&L,&n);
        for(int i=0;i<n;++i)
            scanf("%d",&a[i]);
        ans1();//求所有蚂蚁掉下去的最短时间 
        ans2();//求所有蚂蚁掉下去的最长时间 
    }
    return 0;
}