POJ 1573-Robot Motion(模拟)

标签：

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

Output

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)

题意：有n行m列，t列从北方进入的列数。每一个字母都是它的指令。如果机器人能走出去，输出走出的步数，若在里面转圈，输出第一个为到达圈之前的步数，输出第二个为从开始到转完一次圈的步数。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main()
{
    int n,m,t;
    int i,j;
    char str[30][30];
    char vis[30][30];
    int cnt,loop;
    int flag;
    int x,y;
    while(~scanf("%d %d %d",&n,&m,&t))
    {
        if(n==0&&m==0&&t==0)
        break;
        flag = 0;
        cnt=1;
        loop = 0;

        memset(str,0,sizeof(str));
        memset(vis,0,sizeof(vis));
        for(i = 0; i<n; i++)
        scanf("%s",str[i]);
        x = 0;
        y = t-1;
        while(1)
        {
            if(str[x][y] == 'N'&&!vis[x][y])
            {
                vis[x][y] = cnt;
                x--;
            }
            else if(str[x][y] == 'S' && !vis[x][y])
            {
                vis[x][y] = cnt;
                x++;
            }
            else if(str[x][y] == 'E' && !vis[x][y])
            {
                vis[x][y] = cnt;
                y++;
            }
            else if(str[x][y] == 'W' && !vis[x][y])
            {
                vis[x][y] = cnt;
                y--;
            }
            else if(x < 0 || x >= n|| y<0  || y>=m)
            {
                cnt--;
                break;
            }
            else if(vis[x][y]!=0)
            {
                cnt--;
                loop = cnt-vis[x][y]+1;
                flag = 1;
                break;
            }

            cnt++;
        }
        if(flag)
            printf("%d step(s) before a loop of %d step(s)\n",cnt-loop,loop);
        else
            printf("%d step(s) to exit\n",cnt);
    }

    return 0;
}

POJ 1573-Robot Motion(模拟)

标签：

原文地址：http://blog.csdn.net/u013486414/article/details/42607733