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 = 100;

int dx[] = {-1, 0, 1, 0};//上右下左
int dy[] = {0, 1, 0, -1};

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[0][s-1] = 1;
    vis[0][s-1] = true;

    int x = 0, y = s-1, nx, ny;

    while(true) {
        switch(G[x][y]) {
        case 'N': nx = x + dx[0]; ny = y + dy[0]; break;
        case 'E': nx = x + dx[1]; ny = y + dy[1]; break;
        case 'S': nx = x + dx[2]; ny = y + dy[2]; break;
        case 'W': nx = x + dx[3]; ny = y + dy[3]; break;
        }

        if(nx < 0 || nx >= n || ny < 0 || ny >= m) {
            ex_step = step[x][y];
            return true;
        }

        if(!vis[nx][ny]) {  //exit
            vis[nx][ny] = true;
            step[nx][ny] = step[x][y] + 1;
            x = nx; y = ny;
        }
        else {//发现环
            lo_step = step[x][y] + 1 - step[nx][ny];
            ex_step = step[nx][ny] - 1;
            return false;
        }
    }
}

int main() {
    int s;

    while(scanf("%d%d%d", &n, &m, &s) == 3) {
        if(n == 0 && m == 0 && s == 0) break;

        for(int i=0; i<n; i++) scanf("%s", G[i]);

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

    return 0;
}