思路：

bfs，用状压表示走过的区域，然后和x1，y1，x2，y2构成所有的状态，然后标记一下就可以了

代码：

#pragma GCC optimize(2)#pragma GCC optimize(3)#pragma GCC optimize(4)#include
     
      using namespace std;#define fi first#define se second#define pi acos(-1.0)#define LL long long//#define mp make_pair#define pb push_back#define ls rt<<1, l, m#define rs rt<<1|1, m+1, r#define ULL unsigned LL#define pll pair
      
       #define pii pair
       
        #define piii pair
        
         #define mem(a, b) memset(a, b, sizeof(a))#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);//head bool vis[(1<<16) + 10][5][5][5][5];int dir[4][2] = {
   0, 1, 1, 0, 0, -1, -1, 0};char s[10][10];int n, m;struct node {    int x1, y1, x2, y2, dis, st;};int get(int x, int y) {    return (x-1)*m + y - 1;}int bfs() {    queue
         
           q;    int st = 0;    for (int i = 1; i <= n; i++) {        for (int j = 1; j <= m; j++) {            if(s[i][j] == 'S' || s[i][j] == 'X') st |= 1<