POJ1573 Robot Motion(模拟)

题目链接。

分析：

很简单的一道题，

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <stack>
#include <cmath>
#include <queue>

using namespace std;

const int maxn = ;

int dx[] = {-, , , };//上右下左
int dy[] = {, , , -};

char G[maxn][maxn];
int step[maxn][maxn];
bool vis[maxn][maxn];
int n, m, ex_step, lo_step;

bool solve(int s) {
    memset(vis, false, sizeof(vis));
    step[][s-] = ;
    vis[][s-] = true;

    int x = , y = s-, nx, ny;

    while(true) {
        switch(G[x][y]) {
        case 'N': nx = x + dx[]; ny = y + dy[]; break;
        case 'E': nx = x + dx[]; ny = y + dy[]; break;
        case 'S': nx = x + dx[]; ny = y + dy[]; break;
        case 'W': nx = x + dx[]; ny = y + dy[]; break;
        }

        if(nx <  || nx >= n || ny <  || ny >= m) {
            ex_step = step[x][y];
            return true;
        }

        if(!vis[nx][ny]) {  //exit
            vis[nx][ny] = true;
            step[nx][ny] = step[x][y] + ;
            x = nx; y = ny;
        }
        else {//发现环
            lo_step = step[x][y] +  - step[nx][ny];
            ex_step = step[nx][ny] - ;
            return false;
        }
    }
}

int main() {
    int s;

    while(scanf("%d%d%d", &n, &m, &s) == ) {
        if(n ==  && m ==  && s == ) break;

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

        if(solve(s)) printf("%d step(s) to exit\n", ex_step);
        else printf("%d step(s) before a loop of %d step(s)\n", ex_step, lo_step);
    }

    return ;
}