题目意思

给出一个n * m的地图，起始点是S，终点E，中途#代表障碍物，无法走过。问能不能从起点走到终点。

解题思路

BFS走图题，别用dfs，规模一大就会暴毙。用了bfs的化应该就很简单了，这个题目本身没有难度，就是一个裸的模板题，如果还有不会的康康代码应该就懂了

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(vv) (vv).begin(), (vv).end()
#define endl "\n"
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 500 + 7;
char mp[N][N];
bool vis[N][N];
int dx[] = { 1,-1,0,0 };
int dy[] = { 0,0,1,-1 };

int main() {
    int n, m;
    while (~scanf("%d %d", &n, &m)) {
        memset(vis, 0, sizeof(vis));
        pair<int, int> st, ed;
        for (int i = 0; i < n; ++i) {
            scanf("%s", mp + i);
            for (int j = 0; j < m; ++j)
                if (mp[i][j] == 'S')    st.first = i, st.second = j;
                else if (mp[i][j] == 'E')    ed.first = i, ed.second = j;
        }
        queue<pair<int, int>> q;
        q.push(st);
        int flag = 0;
        while (q.size()) {
            pair<int, int>    x = q.front();    q.pop();
            for (int i = 0; i < 4; ++i) {
                int xx = x.first + dx[i], yy = x.second + dy[i];
                if (xx < 0 or  xx >= n or yy < 0 or yy >= m or mp[xx][yy] == '#' or vis[xx][yy])
                    continue;
                pair<int, int> y = make_pair(xx, yy);
                if (y == ed) {
                    flag = 1;
                    break;
                }
                q.push(y);
                vis[xx][yy] = 1;
            }
            if (flag)    break;
        }
        if (flag)    puts("Yes");
        else    puts("No");
    }
    return 0;
}