ACM题目————Robot Motion

最新推荐文章于 2018-07-27 17:16:50 发布

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原文链接：http://www.cnblogs.com/Asimple/p/5482295.html

本文介绍了一种用于确定机器人在网格中导航至出口所需步骤数或进入循环的算法。通过直接模拟和深度优先搜索两种方法，实现对机器人路径的有效分析。

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Description

A robot has been programmed to follow the instructions in its path. Instructions for the next direction the robot is to move are laid down in a grid. The possible instructions are

N north (up the page)
S south (down the page)
E east (to the right on the page)
W west (to the left on the page)

For example, suppose the robot starts on the north (top) side of Grid 1 and starts south (down). The path the robot follows is shown. The robot goes through 10 instructions in the grid before leaving the grid.

Compare what happens in Grid 2: the robot goes through 3 instructions only once, and then starts a loop through 8 instructions, and never exits.

You are to write a program that determines how long it takes a robot to get out of the grid or how the robot loops around.

Input

There will be one or more grids for robots to navigate. The data for each is in the following form. On the first line are three integers separated by blanks: the number of rows in the grid, the number of columns in the grid, and the number of the column in which the robot enters from the north. The possible entry columns are numbered starting with one at the left. Then come the rows of the direction instructions. Each grid will have at least one and at most 10 rows and columns of instructions. The lines of instructions contain only the characters N, S, E, or W with no blanks. The end of input is indicated by a row containing 0 0 0.

Output

For each grid in the input there is one line of output. Either the robot follows a certain number of instructions and exits the grid on any one the four sides or else the robot follows the instructions on a certain number of locations once, and then the instructions on some number of locations repeatedly. The sample input below corresponds to the two grids above and illustrates the two forms of output. The word "step" is always immediately followed by "(s)" whether or not the number before it is 1.

Sample Input

3 6 5
NEESWE
WWWESS
SNWWWW
4 5 1
SESWE
EESNW
NWEEN
EWSEN
0 0 0

Sample Output

10 step(s) to exit
3 step(s) before a loop of 8 step(s)

看懂题目就好其实不难。

题目大意就是：给出r行c列, 然后从第一行的第i个位置开始走 每个位置有标记E W S N, 分别表示向东, 向西, 向南, 向北 然后, 有两种情况, 一种能走出去, 一种是陷入死循环。要求输出结果！

解法一：（直接模拟）

#include<stdio.h>  
  
char str[100][100];  
  
int main() {  
    int r, c, in;  
    while(scanf("%d %d %d", &r, &c, &in) != EOF) {  
        if(r == 0 && c ==0 && in == 0)  
            break;  
        getchar();  
        for(int i = 0; i < r; i++)  
            gets(str[i]);  
        int tr, tc, loop;  
        int step = 0;  
        int mark = 1;  
        tr = 0; tc = in-1;  
  
        while(1) {  
            if(str[tr][tc] == 'E') {  
                step++;  
                str[tr][tc] = step + '0';  
                if(tc == c-1)  
                    break;  
                else  
                    tc = tc + 1;  
            }  
            else if(str[tr][tc] == 'W') {  
                step++;  
                str[tr][tc] = step + '0';  
                if(tc == 0)  
                    break;  
                else  
                    tc = tc - 1;  
            }  
            else if(str[tr][tc] == 'S') {  
                step++;  
                str[tr][tc] = step + '0';  
                if(tr == r-1)  
                    break;  
                else  
                    tr = tr + 1;  
            }  
            else if(str[tr][tc] == 'N') {  
                step++;  
                str[tr][tc] = step + '0';  
                if(tr == 0)  
                    break;  
                else  
                    tr = tr - 1;  
            }  
            else {  
                mark = 0;  
                loop = str[tr][tc] - '0' - 1;  
                printf("%d step(s) before a loop of %d step(s)\n", loop, step - loop);  
                break;  
            }  
        }  
        if(mark == 1)  
            printf("%d step(s) to exit\n", step);  
    }  
    return 0;  
}


解法二：（DFS）

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
char map[15][15] ;       //用来记录图形
int des[15][15] ;        //用来记录该点所走的步数，以及判断是否构成循环
int a ,b , l , count , f , h;//a,b记录输入的行列数，count记录循环频数，h，循环前的步数
void DFS( int m , int n , int sp )
{
    if( m <0 || m >= a || n < 0 || n >= b )//判断是否出去
    {
        f = 1 ;
        count = sp ;
        return ;
    }
    if( des[m][n] )//判断是否循环
    {
        f = 0 ;
        count = des[m][n];
        h = sp - des[m][n] ;
        return ;
    }
    des[m][n] = sp ; //把步数赋给des
    if( map[m][n] == 'N' )
        DFS( m-1 , n , sp+1) ;
    if( map[m][n] == 'S' )
        DFS( m+1 , n , sp+1 ) ;
    if( map[m][n] == 'E' )
        DFS( m , n+1 , sp+1 ) ;
    if( map[m][n] == 'W' )
        DFS( m , n-1 , sp+1 ) ;
}
int main(  )
{
    while( scanf("%d%d" , &a, &b ) , a || b )
    {
        scanf("%d" , &l ) ;
        memset( des , 0 , sizeof( des ) ) ;
        getchar ( ) ;
        count = 0 ;
        f = 0 ;
        h = 0 ;
        for( int i = 0 ; i < a ; i++ )
        {
            for( int j = 0 ; j < b ; j++ )
                scanf("%c" , &map[i][j] ) ;
            getchar();
        }

        DFS ( 0 ,l-1 , 1 ) ;
        if( f )
        {
            printf("%d step(s) to exit\n" , count-1 ) ;
        }
        else
            printf("%d step(s) before a loop of %d step(s)\n" ,count-1 ,h  ) ;

    }

    return 0;
}