J - New Year and Buggy Bot 暴力枚举

传送门
思路：思路很清晰，简单。通过分析总共会有24种映射情况，所以分别枚举每一种情况，尝试即可。难得是模拟出来，想了好久。

#include<iostream>
#include<algorithm>
using namespace std;
char a[101][101];
string s;
int sx, sy;
int n, m;
void Map(int i)
{
    if(i == 1)
        sx ++;
    if(i == 2)
        sx --;
    if(i == 3)
        sy++;
    if(i == 4)
        sy--;
}
int main()
{
    cin >> n >> m;
    int xt,yt;
    int cnt = 0;
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
        {
            cin >> a[i][j];
            if(a[i][j] == 'S')
                sx = i, sy = j, xt = i, yt = j;
        }
    cin >> s;
    for(int i = 1; i <= 4; i++)
        for(int j = 1; j <= 4; j++)
            for(int k = 1; k <= 4; k++)
                for(int q = 1; q <= 4; q++)
                {
                    if(i != j && i != k && i != q && j != k && j != q && k != q)
                    {
                        sx = xt, sy = yt;
                        for(int h = 0 ;h < s.size();h++)
                        {
                            if(s[h] == '0')
                            {
                                Map(i);
                            }
                            if(s[h] == '1')
                            {
                                Map(j);
                            }
                            if(s[h] == '2')
                            {
                                Map(k);
                            }
                            if(s[h] == '3')
                            {
                                Map(q);
                            }
                            if(sx <= 0 || sy <= 0 || sx > n || sy > m || a[sx][sy] == '#')
                                {break;}
                            if(a[sx][sy] == 'E')
                            {cnt++; break;}
                        }
                    }
                }
                cout << cnt << endl;
    return 0;
}