C++

$\color{gold}{— > 蓝桥杯辅导课题解}$

思路：

BFS

$从起点出发，每次判断一下上下左右四个格子，符合条件则加入队尾，直至到达终点$

$如果无法到达终点，返回-1$

$\color{red}{样例模拟：}$

#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>

#define x first
#define y second

using namespace std;

const int N = 210;
typedef pair<int, int> pii;

int n, m;
char g[N][N]; // 存储迷宫矩阵
int dist[N][N]; // 判重

int bfs(pii start, pii end) {
    queue<pii> q;
    memset(dist, -1, sizeof dist); // 把距离初始化为-1，-1表示不可到达

    dist[start.x][start.y] = 0;
    q.push(start);

    int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};

    while (q.size()) { // 队列不为空
        auto t = q.front();
        q.pop(); // 队头出队，避免死循环

        for (int i = 0; i < 4; i ++) { // 上下左右
            int x = t.x + dx[i], y = t.y + dy[i];
            if (x < 0 || x >= n || y < 0 || y >= m) continue; // 出界
            if (g[x][y] == '#') continue; // 障碍物
            if (dist[x][y] != -1) continue; // 之前已经遍历过

            dist[x][y] = dist[t.x][t.y] + 1;

            if (end == make_pair(x, y)) return dist[x][y];

            q.push({x, y});
        }
    }

    return -1;
}

int main() {
    int T; cin >> T;

    while (T --) {
        cin >> n >> m;
        for (int i = 0; i < n; i ++) cin >> g[i];

        pii start, end; // 找一下起点和终点
        for (int i = 0; i < n; i ++)
            for (int j = 0; j < m; j ++) 
                if (g[i][j] == 'S') start = {i, j};
                else if (g[i][j] == 'E') end = {i, j};

        int distance = bfs(start, end);

        if (distance == -1) puts("oop!");
        else cout << distance << endl;
    }

    return 0;
}