将题目给出的 $n \times m$ 的矩阵扩充成 $3n \times 3m$ 的矩阵，看作一个九宫格。

然后从中间那个格子出发，最后看能不同格子中坐标相同的点是否被经过，如果同一坐标被经过不止一次，那么输出 Yes，否则 No。

下面是代码，只有 70pts:

#include <bits/stdc++.h>
using namespace std;

const int maxn = 4505;
int n, m;
char graph[maxn][maxn];
bool visited[maxn][maxn];
struct Point {
	int x, y;
} start;

int dx[4] = {0, 0, 1, -1};
int dy[4] = {1, -1, 0, 0};
inline void bfs(const Point &start) {
	queue<Point> que;
	que.push(start);
	memset(visited, 0, sizeof(visited));
	visited[start.x][start.y] = true;
	while (!que.empty()) {
		Point u = que.front();
		que.pop();
		for (int i = 0; i < 4; i++) {
			int vx = u.x + dx[i], vy = u.y + dy[i];
			if (vx < 0 || vx > 3 * n || vy < 0 || vy > 3 * m
				|| visited[vx][vy] || graph[vx][vy] == '#') continue;
			que.push(Point{ vx, vy });
			visited[vx][vy] = true;
		}
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			int cnt = 0;
			if (visited[i][j]) cnt++;
			if (visited[i + n][j + m]) cnt++;
			if (visited[i + n][j]) cnt++;
			if (visited[i][j + m]) cnt++;
			if (visited[i + 2 * n][j + 2 * m]) cnt++;
			if (visited[i + 2 * n][j]) cnt++;
			if (visited[i][j + 2 * m]) cnt++;
			if (visited[i + n][j + 2 * m]) cnt++;
			if (visited[i + 2 * n][j + m]) cnt++;
			if (cnt > 1) {
				printf("Yes\n");
				return;
			}
		}
	}
	printf("No\n");
}

int main() {
	while (scanf("%d%d", &n, &m) != EOF) {
		for (int i = n + 1; i <= 2 * n; i++) {
			for (int j = m + 1; j <= 2 * m; j++) {
				scanf(" %c", &graph[i][j]);
				if (graph[i][j] == 'S') {
					start.x = i, start.y = j;
				}
			}
		}
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= m; j++) {
				graph[i][j] = graph[i + n][j + m];
				graph[i + n][j] = graph[i + n][j + m];
				graph[i][j + m] = graph[i + n][j + m];
				graph[i + 2 * n][j] = graph[i + n][j + m];
				graph[i][j + 2 * m] = graph[i + n][j + m];
				graph[i + 2 * n][j + m] = graph[i + n][j + m];
				graph[i + n][j + 2 * m] = graph[i + n][j + m];
				graph[i + 2 * n][j + 2 * m] = graph[i + n][j + m];
			}
		}
		bfs(start);
	}
	return 0;
}