zoukankan      html  css  js  c++  java
  • POJ 2632, Crashing Robots

    模拟类


    Description
    In a modernized warehouse, robots are used to fetch the goods. Careful planning is needed to ensure that the robots reach their destinations without crashing into each other. Of course, all warehouses are rectangular, and all robots occupy a circular floor space with a diameter of 1 meter. Assume there are N robots, numbered from 1 through N. You will get to know the position and orientation of each robot, and all the instructions, which are carefully (and mindlessly) followed by the robots. Instructions are processed in the order they come. No two robots move simultaneously; a robot always completes its move before the next one starts moving.
    A robot crashes with a wall if it attempts to move outside the area of the warehouse, and two robots crash with each other if they ever try to occupy the same spot.

    Input
    The first line of input is K, the number of test cases. Each test case starts with one line consisting of two integers, 1 <= A, B <= 100, giving the size of the warehouse in meters. A is the length in the EW-direction, and B in the NS-direction.
    The second line contains two integers, 1 <= N, M <= 100, denoting the numbers of robots and instructions respectively.
    Then follow N lines with two integers, 1 <= Xi <= A, 1 <= Yi <= B and one letter (N, S, E or W), giving the starting position and direction of each robot, in order from 1 through N. No two robots start at the same position.

     
    Figure 1: The starting positions of the robots in the sample warehouse

    Finally there are M lines, giving the instructions in sequential order.
    An instruction has the following format:
    < robot #> < action> < repeat>
    Where is one of
    L: turn left 90 degrees,
    R: turn right 90 degrees, or
    F: move forward one meter,

    and 1 <= < repeat> <= 100 is the number of times the robot should perform this single move.

     

    Output
    Output one line for each test case:
    Robot i crashes into the wall, if robot i crashes into a wall. (A robot crashes into a wall if Xi = 0, Xi = A + 1, Yi = 0 or Yi = B + 1.)
    Robot i crashes into robot j, if robots i and j crash, and i is the moving robot.
    OK, if no crashing occurs.

    Only the first crash is to be reported.

    Sample Input
    4
    5 4
    2 2
    1 1 E
    5 4 W
    1 F 7
    2 F 7
    5 4
    2 4
    1 1 E
    5 4 W
    1 F 3
    2 F 1
    1 L 1
    1 F 3
    5 4
    2 2
    1 1 E
    5 4 W
    1 L 96
    1 F 2
    5 4
    2 3
    1 1 E
    5 4 W
    1 F 4
    1 L 1
    1 F 20

    Sample Output
    Robot 1 crashes into the wall
    Robot 1 crashes into robot 2
    OK
    Robot 1 crashes into robot 2

    Source
    Nordic 2005


    // POJ2632.cpp : Defines the entry point for the console application.
    //

    #include 
    <iostream>
    using namespace std;

    int main(int argc, char* argv[])
    {
        
    int cases;
        cin 
    >> cases;

        
    int width, height;
        
    int robots, instructions;
        unsigned 
    char rbs[101][3]; // x, y, direction
        unsigned char ins[101][3]; // robot #, action, repeat

        
    int status; //0 ok, 1 crash into wall, 2 crash to robot
        int robnum, crashnum, repeat;
        
    int x, y;
        
    for (int i = 0; i < cases; ++i)
        {
            cin 
    >> width >> height;
            cin 
    >> robots >> instructions;

            
    for (int j = 1; j <= robots; ++j)
            {
                cin 
    >> x >> y >> rbs[j][2];
                rbs[j][
    0= x;
                rbs[j][
    1= y;
            }

            
    for (int j = 1; j <= instructions; ++j)
            {
                cin 
    >> robnum >> ins[j][1>> repeat;
                ins[j][
    0= robnum;
                ins[j][
    2= repeat;
            }

            status 
    = 0;
            
    for (int j = 1; j <= instructions; ++j)
            {
                robnum 
    = ins[j][0];
                repeat 
    = ins[j][2];
                
    if (ins[j][1]=='F')
                {
                    
    switch (rbs[robnum][2])
                    {
                    
    case 'N':
                        x 
    = 0; y = 1;
                        
    break;
                    
    case 'S':
                        x 
    = 0; y = -1;
                        
    break;
                    
    case 'E':
                        x 
    = 1; y = 0;
                        
    break;
                    
    case 'W':
                        x 
    = -1; y = 0;
                        
    break;
                    }
                    
    for (int k = 1; k <= repeat; ++k)
                    {
                        
    if (status != 0break;
                        
    if (rbs[robnum][0+ x * k >= 1 && rbs[robnum][0+ x * k <= width 
                            
    && rbs[robnum][1+ y * k >= 1 && rbs[robnum][1+ y * k <= height)
                        {
                            
    for (int l = 1; l <= robots; ++l)
                                
    if (rbs[robnum][0+ x * k == rbs[l][0&& rbs[robnum][1+ y * k == rbs[l][1])
                                {
                                    status 
    = 2;
                                    crashnum 
    = l;
                                    
    break;
                                };
                        }
                        
    else
                        {
                            status 
    = 1;
                        }
                    };
                    
    if (status == 0)
                    {
                        rbs[robnum][
    0+= x * repeat;
                        rbs[robnum][
    1+= y * repeat;
                    }
                    
    else
                        
    break;
                }
                
    else if (ins[j][1]=='L')
                {
                    
    for (int k = 1; k <= repeat; ++k)
                    {
                        
    switch (rbs[robnum][2])
                        {
                        
    case 'N':
                            rbs[robnum][
    2= 'W';
                            
    break;
                        
    case 'S':
                            rbs[robnum][
    2= 'E';
                            
    break;
                        
    case 'E':
                            rbs[robnum][
    2= 'N';
                            
    break;
                        
    case 'W':
                            rbs[robnum][
    2= 'S';
                            
    break;
                        }
                    }
                }
                
    else
                {
                    
    for (int k = 1; k <= repeat; ++k)
                    {
                        
    switch (rbs[robnum][2])
                        {
                        
    case 'N':
                            rbs[robnum][
    2= 'E';
                            
    break;
                        
    case 'S':
                            rbs[robnum][
    2= 'W';
                            
    break;
                        
    case 'E':
                            rbs[robnum][
    2= 'S';
                            
    break;
                        
    case 'W':
                            rbs[robnum][
    2= 'N';
                            
    break;
                        }
                    }
                }
            }
            
    if (status == 0) cout << "OK\n";
            
    else if (status == 1)
            {
                cout 
    << "Robot " <<robnum <<" crashes into the wall\n";
            }
            
    else if (status == 2)
            {
                cout 
    << "Robot " <<robnum <<" crashes into robot " << crashnum << "\n";
            }
        }

        
    return 0;
    }


  • 相关阅读:
    CF-807B
    CF-807A
    sort()的升降序函数操作
    CF-805D
    CF-805B
    CF-805A
    CF-796C
    CF-796B
    图论学习四之Disjoint set union并查集
    图论学习三之Shortest Path最短路
  • 原文地址:https://www.cnblogs.com/asuran/p/1576047.html
Copyright © 2011-2022 走看看