题解 | #走出迷宫#

BFS做法：

#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
typedef pair<int, int> PII;
const int N = 505;
char map[N][N];
bool st[N][N];
int n, m, flag = 0;
int sx, sy, ex, ey;

void bfs(int x, int y) {
    queue<PII>q;
    q.push({ x, y });
    memset(st, 0, sizeof(st)); // 关键是这行， 有这行就100%对
    st[x][y] = true;

    int dx[] = { 1, 0, -1, 0 }, dy[] = { 0, -1, 0, 1 };
    while (q.size()) {
        PII t = q.front();
        q.pop();
        for (int i = 0; i < 4; i++) {
            int xx = t.first + dx[i], yy = t.second + dy[i];
            if (xx >= 0 && xx < n && yy >= 0 && yy < m && !st[xx][yy] && map[xx][yy] != '#') { // 注意：map[xx][yy]这个地方最好是！=‘#’
                q.push({ xx, yy });
                st[xx][yy] = true;
//                cout << "xx = " << xx << "yy = " << yy << endl;
                if (xx == ex && yy == ey) {
                    flag = 1;
                    return;
                }
            }
        }
    }
}


int main() {

    while (cin >> n >> m) {

        for (int i = 0; i < n; i++)
                for (int j = 0; j < m; j++) {
                    cin >> map[i][j];
                    if (map[i][j] == 'S') sx = i, sy = j;
                    if (map[i][j] == 'E') ex = i, ey = j;
                }
//        cout << "2ex = " << ex << " "<< "2ey = " << ey << endl;
        bfs(sx, sy);
            if (flag == 1) cout << "Yes" << endl;
            else cout << "No" << endl;
            flag = 0;

    }



    return 0;
}