zoukankan      html  css  js  c++  java
  • Crashing Robots 分类: POJ 2015-06-29 11:44 10人阅读 评论(0) 收藏

    Crashing Robots
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 8340   Accepted: 3607

    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
    模拟题
    
    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <cmath>
    #include <string>
    #include <stack>
    #include <queue>
    #include <vector>
    #include <algorithm>
    using namespace std;
    
    const int Max=1100000;
    
    struct node
    {
        int dir;
        int x;
        int y;
    } Robot[110];
    
    struct INS
    {
        int num;
        int action;
        int repeat;
    } Order[110];
    
    bool Map[110][110];
    
    int Dir[4][2]= {{1,0},{0,-1},{-1,0},{0,1}};
    
    int A,B;
    
    int n,m;
    
    int Handle(int s)
    {
        if(s=='S'||s=='R')
        {
            return 1;
        }
        if(s=='N')
        {
            return 3;
        }
        if(s=='E'||s=='F')
        {
            return 0;
        }
        if(s=='W')
        {
            return 2;
        }
        if(s=='L')
            return -1;
        return 0;
    }
    bool Mon()
    {
        for(int i=0; i<m; i++)
        {
            if(Order[i].action==1)
            {
                while(Order[i].repeat--)
                {
                    Robot[Order[i].num].dir++;
                    if(Robot[Order[i].num].dir==4)
                    {
                        Robot[Order[i].num].dir=0;
                    }
                }
            }
            else if(Order[i].action==-1)
            {
                while(Order[i].repeat--)
                {
                    Robot[Order[i].num].dir--;
                    if(Robot[Order[i].num].dir==-1)
                    {
                        Robot[Order[i].num].dir=3;
                    }
                }
            }
            else if(Order[i].action==0)
            {
                while(Order[i].repeat--)
                {
                    Map[Robot[Order[i].num].x][Robot[Order[i].num].y]=false;
                    Robot[Order[i].num].x+=Dir[Robot[Order[i].num].dir][0];
                    Robot[Order[i].num].y+=Dir[Robot[Order[i].num].dir][1];
                    if(Robot[Order[i].num].x==0||Robot[Order[i].num].x==A+1||Robot[Order[i].num].y==0||Robot[Order[i].num].y==B+1)
                    {
                        printf("Robot %d crashes into the wall
    ",Order[i].num);
                        return false;
                    }
                    else if(Map[Robot[Order[i].num].x][Robot[Order[i].num].y])
                    {
                        for(int j=1; j<=n; j++)
                        {
                            if(j!=Order[i].num&&Robot[j].x==Robot[Order[i].num].x&&Robot[j].y==Robot[Order[i].num].y)
                            {
                                printf("Robot %d crashes into robot %d
    ",Order[i].num,j);
                                return false;
                            }
                        }
                    }
                    else
                    {
                        Map[Robot[Order[i].num].x][Robot[Order[i].num].y]=true;
                    }
                }
    
            }
        }
        return true;
    }
    int main()
    {
        int T;
        char s;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d %d",&A,&B);
            scanf("%d %d",&n,&m);
            memset(Map,false,sizeof(Map));
            for(int i=1; i<=n; i++)
            {
                scanf("%d %d %c",&Robot[i].x,&Robot[i].y,&s);
                Map[Robot[i].x][Robot[i].y]=true;
                Robot[i].dir=Handle(s);
            }
            for(int i=0; i<m; i++)
            {
                scanf("%d %c %d",&Order[i].num,&s,&Order[i].repeat);
                Order[i].action=Handle(s);
            }
            if(Mon())
            {
                printf("OK
    ");
            }
        }
        return 0;
    }
    


    版权声明:本文为博主原创文章,未经博主允许不得转载。

  • 相关阅读:
    7、单向一对多的关联关系(1的一方有n的一方的集合属性,n的一方却没有1的一方的引用)
    6、JPA_映射单向多对一的关联关系(n的一方有1的引用,1的一方没有n的集合属性)
    解决ubuntu的screen已经处于Attached状态,无法再打开窗口
    关于.ssh出错,无法从远程git仓库拉代码
    给程序添加git commit信息
    ubuntu服务器常用命令
    uint128_t 添加 c++ 重载类型强制转换
    Visual Studio 查看宏展开
    EOS dice移到1.8版本的修改汇总
    ubuntu 添加字体
  • 原文地址:https://www.cnblogs.com/juechen/p/4721971.html
Copyright © 2011-2022 走看看