ACM: 简单题 uva 10881

Problem D Piotr's Ants

 

"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 T seconds, print "Fell off" for that ant. Print an empty line after each test case.

Sample Input

2

10 1 4

1 R

5 R

3 L

10 R

10 2 3

4 R

5 L

8 R

Sample Output

Case #1:

2 Turning

6 R

2 Turning

Fell off

Case #2:

3 L

6 R

10 R

题意: 有n只蚂蚁在一根长为L厘米的木棍上爬, 方向只有左右, 每1s行走1厘米, 蚂蚁相碰时会同时转向.

         输出经过Ts后全部蚂蚁位置.

解题思路:

        1. 其实只考虑初始和结束位置即可.  例如2只蚂蚁, 1=(1,R), 2=(3,L), 经过2s后结果是(3,R), (1,L);

            显然我们确定结束位置谁是谁即可. 你可以发现, 其实蚂蚁从左到右的相对位置是不变的.

代码:

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
#define MAX 10005

struct node
{
    int num, pos, dir;
    node(int n, int p, int d)
    {
        num = n, pos = p, dir = d;
    }
    bool operator <(const node &x) const
    {
        return pos < x.pos;
    }
};

char dire[3][10] = {"L", "Turning", "R"};

int L, T, n;
int order[MAX];

int main()
{
//    freopen("input.txt", "r", stdin);
    int num = 1;
    int caseNum;
    scanf("%d", &caseNum);
    while(caseNum--)
    {
        vector first, end;
        scanf("%d %d %d", &L, &T, &n);
        int pos, d;
        char str[2];
        for(int i = 0; i < n; ++i)
        {
            scanf("%d %s", &pos, str);
            if(str[0] == 'L') d = -1;
            else d = 1;
            first.push_back( node(i, pos, d) );
            end.push_back( node(0, pos+T*d, d) );
        }
       
        sort(first.begin(), first.end());
        for(int i = 0; i < n; ++i)
            order[ first[i].num ] = i;
           
        sort(end.begin(), end.end());
        for(int i = 0; i < n-1; ++i)
        {
            if(end[i].pos == end[i+1].pos)
                end[i].dir = end[i+1].dir = 0;
        }
       
        printf("Case #%d:\n", num++);
        for(int i = 0; i < n; ++i)
        {
            int temp = order[i];
            if(end[temp].pos < 0 || end[temp].pos > L)
                printf("Fell off\n");
            else
                printf("%d %s\n", end[temp].pos, dire[end[temp].dir+1]);
        }
        printf("\n");
    }
    return 0;
}