codeforces793 B. Igor and his way to work (dfs)

题目链接：codeforces793 B. Igor and his way to work (dfs)

求从起点到终点转方向不超过两次是否有解，，好水啊，感觉自己代码好搓。。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<cmath>
#include<queue>
#include<limits.h>
#define CLR(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long ll;
const int N = 1005;
int n, m;
int sx, sy, ex, ey;int flag;
char s[N][N];
int vis[N][N][5][5];
void dfs(int x, int y, int d, int turn) {//上下：d = 1，左右：d = 2
    if(turn > 2 || vis[x][y][d][turn]) return;
    if(x == ex && y == ey) {
        flag = true;
        return;
    }
    vis[x][y][d][turn] = 1;
    if(x-1 >= 0 && s[x-1][y] != '*' && !vis[x-1][y][1][turn+(d==2)]) {
        dfs(x-1, y, 1, turn + (d==2));
    }
    if(x+1 < n && s[x+1][y] != '*' && !vis[x+1][y][1][turn+(d==2)]) {
        dfs(x+1, y, 1, turn + (d==2));
    }
    if(y-1 >= 0 && s[x][y-1] != '*' && !vis[x][y-1][2][turn+(d==1)]) {
        dfs(x, y-1, 2, turn + (d==1));
    }
    if(y+1 < m && s[x][y+1] != '*' && !vis[x][y+1][2][turn+(d==1)]) {
        dfs(x, y+1, 2, turn + (d==1));
    }
}
int main() {
    int i, j;
    scanf("%d%d ", &n, &m);
    for(i = 0; i < n; ++i) {
        scanf("%s", s[i]);
        for(j = 0; j < m; ++j) {
            if(s[i][j] == 'S') {sx = i; sy = j;}
            else if(s[i][j] == 'T') {ex = i; ey = j;}
        }
    }
    flag = 0;
    CLR(vis, 0);
    dfs(sx, sy, 0, 0);
    if(flag) puts("YES");
    else puts("NO");
    return 0;
}