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bfs：求最短路径的长度

题目：迷宫的最短路径
给定一个大小为N x M的迷宫。迷宫由通道和墙壁组成。每一步可以向邻接的上下左右四格的通道
移动。请求出从起点到终点所需的最小步数

#S######.#
......#..#
.#.##.##.#
.#........
##.##.####
....#....#
.#######.#
.#######.#
....#.....
.####.###.
....#...G#

输出 22

#include <cstdio>
#include <queue>

using namespace std;

const int inf = 0x3fffffff;
const int maxn = 105;

char g[maxn][maxn];
int sx, sy;
int gx, gy;
int n;
int m;
int d[maxn][maxn];

int bfs() {
    queue<pair<int, int> > que;

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            d[i][j] = inf;
        }
    }

    que.push(make_pair(sx, sy));
    d[sx][sy] = 0;

    while (que.size()) {
        pair<int, int> p = que.front();
        que.pop();

        if (p.first == gx && p.second == gy) {
            break;
        }

        int dx[4] = { 1, 0, -1, 0 };
        int dy[4] = { 0, 1, 0, -1 };

        for (int i = 0; i < 4; i++) {
            int nx = p.first + dx[i];
            int ny = p.second + dy[i];
            if (0 <= nx && nx < n && 0 <= ny && ny < m && g[nx][ny] != ‘#‘ && d[nx][ny] == inf) {
                que.push(make_pair(nx, ny));
                d[nx][ny] = d[p.first][p.second] + 1;
            }
        }
    }
    return d[gx][gy];
}

void solve() {
    int res = bfs();
    printf("%d\n", res);
}

int main(void) {
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            scanf("%c", &g[i][j]); // 如果超时可以考虑用 scanf("%s", g[i]);
            if (g[i][j] == ‘S‘) {
                sx = i;
                sy = j;
            } else if (g[i][j] == ‘G‘) {
                gx = i;
                gy = j;
            }
        }
        getchar();
    }
    solve();

    return 0;
}