poj1573--Robot Motion

题意： 给你一个nXm的矩阵 里面放了ESWN四个字母分表代表东南西北然后给你一个位置 代表第一列（最上面一列）的第k个位置，然后要求求出机器人根据东南西北的指令最终会在第几步走出地图或者在第几步进入一个一共几步的循环

---

原题：

Robot Motion

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13477		Accepted: 6509

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)

很简单的模拟 看代码应该就能看懂 只要模拟机器人一直走就好，开一个状态数组记录位置是否走过 

题很水不说太多 直接上代码：

#include <iostream>
#include <string.h>

using namespace std;

int main()
{
    int n,m,po,pox,poy,step;
    bool flag;
    char rmap[100][100];
    bool vis[100][100];
    while(cin>>n>>m>>po&&(n!=0||m!=0||po!=0))
    {
        memset(vis,0,sizeof(vis));
        flag = false;
        pox = po-1;
        poy = 0;
        step = 0;
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++)
                cin>>rmap[i][j];
        while(pox>=0&&poy>=0&&pox<m&&poy<n){
            if(!vis[poy][pox]){
                vis[poy][pox] = true;
                step++;
                if(rmap[poy][pox]=='W'){
                    pox--;
                }else if(rmap[poy][pox]=='E'){
                    pox++;
                }else if(rmap[poy][pox]=='S'){
                    poy++;
                }else if(rmap[poy][pox]=='N'){
                    poy--;
                }
            }else{
                vis[poy][pox] = false;
                flag = true;
                break;
            }
        }
        if(flag){
            pox = po-1;
            poy = 0;
            int count = 0;
            while(vis[poy][pox]){
                count++;
                if(rmap[poy][pox]=='W'){
                    pox--;
                }else if(rmap[poy][pox]=='E'){
                    pox++;
                }else if(rmap[poy][pox]=='S'){
                    poy++;
                }else if(rmap[poy][pox]=='N'){
                    poy--;
                }
            }
            cout<<count<<" step(s) before a loop of "<<step-count<<" step(s)"<<endl;
        }else{
            cout<<step<<" step(s) to exit"<<endl;
        }
    }
    return 0;
}