Crashing Robots(POJ--2632-优快云博客

本文链接：https://blog.youkuaiyun.com/zhang_di233/article/details/47156677

本文介绍了一个关于机器人路径规划的竞赛题目，重点在于如何避免多个机器人在执行指令时发生碰撞。文章详细描述了输入输出格式及样例，适用于对机器人路径规划、碰撞检测感兴趣的读者。

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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.

题意：有t组数据，输入a和b，表示房间的东西长度和南北长度，输入n和c，表示n个机器人和c条命令，接下来n行分别输入机器人的初始位置和初始方向，接下来的c行输入机器人的编号，要行使的走向，行使该走向几次，如果机器人在行走过程中碰到墙或机器人就要输出来发生的碰撞错误且只输出第一次的错误，若整个行使过程都没发生碰撞就输出OK。

Sample Input

Sample Output

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

#include <cstdio>
#include <cstring>
using namespace std;
int dx[]= {0,1,0,-1};                                  //顺时针的方向数组
int dy[]= {1,0,-1,0};
int a,b,n;
struct node
{
    int x,y;
    int dir;
} st[110];
int solve(int nn)
{
    if(st[nn].x<1||st[nn].x>a||st[nn].y<1||st[nn].y>b)
    {
        printf("Robot %d crashes into the wall\n",nn);
        return 0;
    }
    for(int i=1; i<=n; i++)
    {
        if(i==nn)
            continue;
        if(st[i].x==st[nn].x&&st[i].y==st[nn].y)
        {
            printf("Robot %d crashes into robot %d\n",nn,i);
            return 0;
        }
    }
    return 1;
}
int pro(int nn,char *cc)
{
    if(cc[0]=='L')
    {
        st[nn].dir--;
        st[nn].dir=st[nn].dir==-1?3:st[nn].dir;
        return 1;
    }
    else if(cc[0]=='R')
    {
        st[nn].dir++;
        st[nn].dir=st[nn].dir%4;
        return 1;
    }
    else if(cc[0]=='F')
    {
        st[nn].x+=dx[st[nn].dir];
        st[nn].y+=dy[st[nn].dir];
        return solve(nn);
    }
}
int main()
{
    //freopen("oo.text","r",stdin);
    int t,com;
    char direction[3];
    while(~scanf("%d",&t))
    {
        while(t--)
        {
            int flag=0;
            scanf("%d %d",&a,&b);
            scanf("%d %d",&n,&com);
            for(int i=1; i<=n; i++)
            {
                scanf("%d %d %s",&st[i].x,&st[i].y,direction);
                if(direction[0]=='N')
                    st[i].dir=0;
                else if(direction[0]=='E')
                    st[i].dir=1;
                else if(direction[0]=='S')
                    st[i].dir=2;
                else if(direction[0]=='W')
                    st[i].dir=3;
            }
            int num,rep;
            char cc[3];
            for(int i=0; i<com; i++)
            {
                scanf("%d %s %d",&num,cc,&rep);
                if(flag)                                          //如果已经发生了碰撞则只接着输入即可
                    continue;
                for(int j=0; j<rep; j++)
                    if(!pro(num,cc))
                    {
                        flag=1;
                        break;
                    }
            }
            if(!flag)
                printf("OK\n");
        }
    }
    return 0;
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