10881 - Piotr's Ants UVA

最新推荐文章于 2022-10-08 07:46:03 发布

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最新推荐文章于 2022-10-08 07:46:03 发布

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本文介绍了一个关于蚂蚁在杆上行走并相互碰撞的模拟问题。重点在于如何通过编程来预测蚂蚁们在给定时间后的最终位置及状态，包括继续行走、转向或是跌落杆外。

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Piotr's Ants
Time Limit: 2 seconds

"One thing is for certain: there is no stopping them;
the ants will soon be here. And I, for one, welcome our
new insect overlords."

Kent Brockman

Piotr likes playing with ants. He has n of them on a horizontal pole L cm long. Each ant is facing either left or right and walks at a constant speed of 1 cm/s. When two ants bump into each other, they both turn around (instantaneously) and start walking in opposite directions. Piotr knows where each of the ants starts and which direction it is facing and wants to calculate where the ants will end up T seconds from now.

Input
The first line of input gives the number of cases, N. N test cases follow. Each one starts with a line containing 3 integers: L , T and n (0 <= n <= 10000). The next n lines give the locations of the n ants (measured in cm from the left end of the pole) and the direction they are facing (L or R).

Output
For each test case, output one line containing "Case #x:" followed by n lines describing the locations and directions of the n ants in the same format and order as in the input. If two or more ants are at the same location, print "Turning" instead of "L" or "R" for their direction. If an ant falls off the pole before Tseconds, print "Fell off" for that ant. Print an empty line after each test case.

Sample Input	Sample Output
2 10 1 4 1 R 5 R 3 L 10 R 10 2 3 4 R 5 L 8 R	Case #1: 2 Turning 6 R 2 Turning Fell off Case #2: 3 L 6 R 10 R

对于这一些运动的蚂蚁，当他们因碰撞而掉头时，看上去和两只蚂蚁对穿而过没有任何区别；

例如三只蚂蚁（1，R）（3，L)（4，L)两秒过后三只蚂蚁分别是（3，R)（1，L)（2，L)

但并不是一一对应的关系！！！！！！！！！

另外本题的关键点是所有蚂蚁的相对位置是不变的！！！！！！！！！

对before和after数组进行排序，那么这两个数组是一一对应的；因为

输入顺序与排序后的before已经不一样所以需要用数组order保存输入顺序

以下是ac代码：

#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <string.h>
#include <stdlib.h>
using namespace std;
const int maxn = 10;
struct ant
{
    int index;
    int pos;
    int dir;
}before[10000+maxn],after[10000+maxn];
int order[10000+maxn];
bool cmp(ant a1,ant a2)
{
    return a1.pos<a2.pos;
}
int main()
{
    int N;
    scanf("%d",&N);
    int i,j;
    for(i = 0;i<N;++i)
    {
        int L,T,n;
        scanf("%d %d %d",&L,&T,&n);
        printf("Case #%d:\n",i+1);
        for(j = 0;j<n;++j)
        {
            char ch;
            scanf("%d %c",&before[j].pos,&ch);
            if(ch == 'R')
            before[j].dir = 1;
            else
            before[j].dir = -1;
            before[j].index = j;
            after[j].index = 0;
            after[j].pos = before[j].pos + T*before[j].dir;
            after[j].dir  = before[j].dir;
        }
        sort(before,before+n,cmp);

        for(j = 0;j<n;++j)
        order[before[j].index] = j;

        sort(after,after+n,cmp);
        for(j = 0;j<n-1;++j)
        if(after[j].pos == after[j+1].pos)
          after[j].dir = after[j+1].dir = 0;

         for(j = 0;j<n;++j)
          {
              int a = order[j];
              if(after[a].pos<0||after[a].pos>L)
              {
              printf("Fell off\n");
              continue;
              }
              if(after[a].dir == 1)
              printf("%d R\n",after[a].pos);
              else
              if(after[a].dir == -1)
              printf("%d L\n",after[a].pos);
              else
              printf("%d Turning\n",after[a].pos);
          }
          printf("\n");
    }
    return 0;
}

注：数组before或after可以用（ant）{index,pos,dir}来进行初始化