poj2632--Crashing Robots

最新推荐文章于 2018-02-05 22:13:00 发布

刀刀狗0102

最新推荐文章于 2018-02-05 22:13:00 发布

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本文链接：https://blog.youkuaiyun.com/winddreams/article/details/26450695

本文介绍了一个关于机器人在仓库中移动并避免碰撞的问题。通过模拟不同指令下机器人的移动路径，确保它们不会相互碰撞或撞到墙壁。文章提供了一种解决方案，使用结构体来跟踪每个机器人的位置和状态。

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Crashing Robots

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7383		Accepted: 3240

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

Sample Output

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

Source

Nordic 2005

模拟题，只是比较多，我用了两个结构体，不知道还有没有好的办法

1 转向

2 向当前方向走num步，在这之中会有碰到其他的机器人，或是在最后撞墙。

3 只要在一步操作后出现问题，以后的操作都不用判断。

#include <stdio.h>
struct node
{
    int flag ;
    int n ;
    char s ;
} p[102][102];
struct no
{
    int x , y ;
} q[102];
int main()
{
    int t ;
    char str[] = "SENW" ;
    scanf("%d", &t);
    while(t--)
    {
        int a , b , i , j  ;
        scanf("%d %d", &b, &a);
        for(i = 1 ; i <= a ; i++)
            for(j = 1 ; j <= b ; j++)
            {
                p[i][j].flag = 0 ;
            }
        int n , m , num ;
        char s ;
        scanf("%d %d", &n, &m);
        for(i = 1 ; i <= n ; i++)
        {
            int x , y ;
            scanf("%d %d%*c%c", &y, &x, &s);
            q[i].x = x ;
            q[i].y = y ;
            p[x][y].flag = 1 ;
            p[x][y].n = i ;
            p[x][y].s = s ;
        }
        int flag = 1 ;
        while(m--)
        {
            scanf("%d%*c%c %d", &n, &s, &num);
            if(flag)
            {
                if(s == 'L')
                {
                    num %= 4 ;
                    j = 0 ;
                    while(1)
                    {
                        if( p[ q[n].x ][ q[n].y ].s == str[j] )
                            break;
                        else
                            j++ ;
                    }
                    j += num ;
                    if( j > 3)
                        j -= 4 ;
                    p[ q[n].x ][ q[n].y ].s = str[j] ;
                }
                else if(s == 'R')
                {
                    num %= 4 ;
                    j = 0 ;
                    while(1)
                    {
                        if( p[ q[n].x ][ q[n].y ].s == str[j] )
                            break;
                        else
                            j++ ;
                    }
                    j -= num ;
                    if( j < 0)
                        j += 4 ;
                    p[ q[n].x ][ q[n].y ].s = str[j] ;
                }
                else
                {
                    int k = 0 ;
                    while(1)
                    {
                        if( p[ q[n].x ][ q[n].y ].s == str[k] )
                            break;
                        else
                            k++ ;
                    }
                    if(k == 0)
                    {
                        for(i = q[n].x-1 ; i >= q[n].x - num ; i--)
                        {
                            if(i == 0)
                            {
                                printf("Robot %d crashes into the wall\n", n);
                                flag = 0 ;
                                break;
                            }
                            else if( p[i][ q[n].y ].flag )
                            {
                                printf("Robot %d crashes into robot %d\n", n, p[i][ q[n].y ].n ) ;
                                flag = 0 ;
                                break;
                            }
                        }
                        if(flag)
                        {
                            p[ q[n].x - num ][ q[n].y ] = p[ q[n].x ][ q[n].y ] ;
                            p[ q[n].x ][ q[n].y ].flag = 0 ;
                            q[n].x = q[n].x - num ;
                        }
                    }
                    else if(k == 1)
                    {
                        for(i = q[n].y+1 ; i <= q[n].y + num ; i++)
                        {
                            if(i == b+1)
                            {
                                printf("Robot %d crashes into the wall\n", n);
                                flag = 0 ;
                                break;
                            }
                            else if( p[ q[n].x ][i].flag )
                            {
                                printf("Robot %d crashes into robot %d\n", n, p[ q[n].x ][i].n ) ;
                                flag = 0 ;
                                break;
                            }
                        }
                        if(flag)
                        {
                            p[ q[n].x  ][ q[n].y + num ] = p[ q[n].x ][ q[n].y ] ;
                            p[ q[n].x ][ q[n].y ].flag = 0 ;
                            q[n].y = q[n].y + num ;
                        }
                    }
                    else if(k == 2)
                    {
                        for(i = q[n].x+1 ; i <= q[n].x + num ; i++)
                        {
                            if(i == a+1)
                            {
                                printf("Robot %d crashes into the wall\n", n);
                                flag = 0 ;
                                break;
                            }
                            else if( p[i][ q[n].y ].flag )
                            {
                                printf("Robot %d crashes into robot %d\n", n, p[i][ q[n].y ].n ) ;
                                flag = 0 ;
                                break;
                            }
                        }
                        if(flag)
                        {
                            p[ q[n].x + num ][ q[n].y ] = p[ q[n].x ][ q[n].y ] ;
                            p[ q[n].x ][ q[n].y ].flag = 0 ;
                            q[n].x = q[n].x + num ;
                        }
                    }
                    else if(k == 3)
                    {
                        for(i = q[n].y-1 ; i >= q[n].y - num ; i--)
                        {
                            if(i == 0)
                            {
                                printf("Robot %d crashes into the wall\n", n);
                                flag = 0 ;
                                break;
                            }
                            else if( p[ q[n].x ][i].flag )
                            {
                                printf("Robot %d crashes into robot %d\n", n, p[ q[n].x ][i].n ) ;
                                flag = 0 ;
                                break;
                            }
                        }
                        if(flag)
                        {
                            p[ q[n].x ][ q[n].y - num ] = p[ q[n].x ][ q[n].y ] ;
                            p[ q[n].x ][ q[n].y ].flag = 0 ;
                            q[n].y = q[n].y - num ;
                        }
                    }
                }
            }
        }
        if(flag)
            printf("OK\n");
    }
}