CF586D. Phillip and Trains

Posted 2020-09-14 BBBob

 1 /*
 2  CF586D. Phillip and Trains
 3  http://codeforces.com/problemset/problem/586/D
 4  搜索
 5  */
 6 #include<cstdio>
 7 #include<algorithm>
 8 #include<string.h>
 9 using namespace std;
10 const int Nmax=110;
11 char map[4][Nmax];
12 int fail[4][Nmax];
13 int n,k;
14 int t,ans;
15 
16 int dfs(int step,int pos)
17 {
18     if(step==n-1 && map[pos][step+1]==‘.‘)
19     {
20         ans=1;
21         return 1;
22     }
23     if(step>=n)
24     {
25         ans=1;
26         return 1;
27     }
28     if(ans)
29         return 1;
30     if(fail[pos][step])
31         return 0;
32     
33     
34     if(map[pos][step]>=‘A‘ && map[pos][step]<=‘Z‘)
35     {
36         fail[pos][step]=1;
37         return 0;
38     }
39     for(int k=pos-1;k<=pos+1;k++)
40     {
41         if(k>=1 && k<=3 && map[pos][step+1]==‘.‘ &&map[k][step+1]==‘.‘ && !fail[k][step+1] && !fail[k][step+2] && !fail[k][step+3])
42         {
43             if(map[k][step+2]==‘.‘ && map[k][step+3]==‘.‘)
44             {
45                 if(dfs(step+3,k))
46                 {
47                     ans=1;
48                     return 1;
49                 }
50                 else
51                 {
52                     fail[k][step+3]=1;
53                 }
54             }
55             
56         }
57     }
58     fail[pos][step]=1;
59     return 0;
60 }
61 
62 int main()
63 {
64     scanf("%d",&t);
65     while(t--)
66     {
67         scanf("%d%d",&n,&k);
68         for(int i=1;i<=3;i++)
69             for(int j=1;j<=n+10;j++)
70                 fail[i][j]=0;
71         getchar();
72         gets(map[1]+1);
73         gets(map[2]+1);
74         gets(map[3]+1);
75         for(int i=1;i<=3;i++)
76             for(int j=n+1;j<=n+10;j++)
77                 map[i][j]=‘.‘;    
78         ans=0;
79         int pos=0,start;
80         for(int i=1;i<=3;i++)
81             if(map[i][1]==‘s‘)
82                 pos=i;
83         
84         if(dfs(1,pos))
85             printf("YES\n");
86         else
87             printf("NO\n");
88     }
89     return 0;
90 }