poj 1573 Robot Motion【模拟题 写个while循环一直到机器人跳出来】

                                                                                                               Robot Motion

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11462		Accepted: 5558

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)

这道题目好像当初大一的时候，ZhaoPeng前辈就挂出过这样的题目让我们来做，记得那时候编程能力很烂，思维也不好，写的代码也很复杂。

那时候，就这道题我记得我做了一个晚上，最后写出来了，样例运行通过，但是提交是错的，印象里可能是RE。今天碰巧再次遇到了，拿过了

写了一下，1A，代码和之前的那份相比也变得非常的简短了！无论自己成长的快与慢，可以肯定的是自己确实成长了！

代码：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <ctype.h>
#include <math.h>
#include <iostream>
#include <string>
#include <algorithm>

using namespace std;

char g[20][20];
bool vis[20][20];
unsigned int cnt[20][20];
int n, m, st;

bool ok(int x, int y)
{
    if(x>=1 && x<=n && y>=1 && y<=m) return true;
    else return false;
}

int main()
{
    int i, j, k;

    while(scanf("%d %d %d%*c", &n, &m, &st))
    {
        if(n==0 && m==0 && st==0) break;

        for(i=1; i<=n; i++)
            scanf("%s", g[i]+1);

        memset(vis, false, sizeof(vis));//
        memset(cnt, 0, sizeof(cnt));

        int pace=0;
        int x=1; int y=st;
        bool flag=false;
        int ans=0;

        while( vis[x][y]==false ){
            cnt[x][y]=pace++; vis[x][y]=true;
            if(ok(x,y)){
                if(g[x][y]=='W'){ x=x; y=y-1; }
                else if(g[x][y]=='E') { x=x; y=y+1; }
                else if(g[x][y]=='S') { x=x+1; y=y; }
                else {x=x-1; y=y; }
            }
            else{
                flag=true;//如果已经走出来
                ans=cnt[x][y];
                break; //跳出循环
            }
        }
        if(flag){
            printf("%d step(s) to exit\n", ans);
        }else{
            printf("%d step(s) before a loop of %d step(s)\n", cnt[x][y], pace-cnt[x][y] );
        }
    }
    return 0;
}