基础深搜题
An escape
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 227 Accepted Submission(s): 56
Problem Description
You are now in a maze. You mark all the blocks you've visited by '@',
so when you see a wall '#' or a visited block '@' in front of you, you
will make a right turn. Otherwise, that means you don't have a wall or a
visited block in front, you'll go forward. When you reach the door 'D', congratulations!
####
#Y@#
##D#
####
Look at the maze above, you are now in 'Y', facing left, and seeing a wall in
front of you. You turn right, a wall again; turn right again, visited block;
turn right once again, still a wall. After three continuous turnings, you realize
the rest time of your life will be making turnings.
Input
The first line is T(T<=20), then T cases follow.
Each case has two numbers n and m(4<=n,m <= 20), the boundary of the maze will
always be '#', in the maze, there will
be exactly one 'Y', one 'D'. Normal blocks are marked with '.'.
At first you are facing left.
Output
"YES" if you can go out of the maze(reach 'D'). "NO" otherwise.
Sample Input
2
4 4
####
#.Y#
##D#
####
4 4
####
#.Y#
#D##
####
Sample Output
NO
YES
Author
MadFroG
自己写的超长代码如下：
1
#include<iostream>2
#include<stdio.h>3
using namespace std;4
char a[21][21];5
void c(int x,int y,int z)//z表示转的方向 16
{7
int p=0;8
if(z==1)9
{10
if(a[x][y-1]=='D')11
{12
printf("YES\n");13
p++;14
}15
16
else if(a[x][y-1]=='.')17
{18
a[x][y]='@';19
p++;20
z=1;21
c(x,y-1,z);22
}23
else if(a[x-1][y]=='D')24
{ printf("YES\n");25
p++;}26
else if(a[x-1][y]=='.')27
{28
a[x][y]='@';29
z=2;30
p++;31
c(x-1,y,z);32
}33
else if(a[x][y+1]=='D')34
{ printf("YES\n");35
p++;}36
else if(a[x][y+1]=='.')37
{38
a[x][y]='@';39
z=3;40
p++;41
c(x,y+1,z);42
}43
else if(a[x+1][y]=='D')44
{ printf("YES\n");45
p++;}46
else if(a[x+1][y]=='.')47
{48
a[x][y]='@';49
z=3;50
p++;51
c(x+1,y,z);52
}53
}54
else if(z==2)55
{56
if(a[x-1][y]=='D')57
{ printf("YES\n");58
p++;}59
else if(a[x-1][y]=='.')60
{61
a[x][y]='@';62
z=2;63
p++;64
c(x-1,y,z);65
}66
else if(a[x][y+1]=='D')67
{ printf("YES\n");68
p++;}69
else if(a[x][y+1]=='.')70
{71
a[x][y]='@';72
z=3;73
p++;74
c(x,y+1,z);75
}76
else if(a[x+1][y]=='D')77
{ printf("YES\n");78
p++;}79
else if(a[x+1][y]=='.')80
{81
a[x][y]='@';82
z=3;83
p++;84
c(x+1,y,z);85
}86
else if(a[x][y-1]=='D')87
{ printf("YES\n");88
p++;}89
else if(a[x][y-1]=='.')90
{91
a[x][y]='@';92
p++;93
z=1;94
c(x,y-1,z);95
}96
}97
else if(z==3)98
{99
if(a[x][y+1]=='D')100
{ printf("YES\n");101
p++;}102
else if(a[x][y+1]=='.')103
{104
a[x][y]='@';105
z=3;106
p++;107
c(x,y+1,z);108
}109
else if(a[x+1][y]=='D')110
{ printf("YES\n");111
p++;}112
else if(a[x+1][y]=='.')113
{114
a[x][y]='@';115
z=3;116
p++;117
c(x+1,y,z);118
}119
else if(a[x][y-1]=='D')120
{ printf("YES\n");121
p++;}122
else if(a[x][y-1]=='.')123
{124
a[x][y]='@';125
p++;126
z=1;127
c(x,y-1,z);128
}129
else if(a[x-1][y]=='D')130
{ printf("YES\n");131
p++;}132
else if(a[x-1][y]=='.')133
{134
a[x][y]='@';135
z=2;136
p++;137
c(x-1,y,z);138
}139
}140
else if(z==4)141
{142
if(a[x+1][y]=='D')143
{ printf("YES\n");144
p++;}145
else if(a[x+1][y]=='.')146
{147
a[x][y]='@';148
z=3;149
p++;150
c(x+1,y,z);151
}152
else if(a[x][y-1]=='D')153
{ printf("YES\n");154
p++;}155
else if(a[x][y-1]=='.')156
{157
a[x][y]='@';158
p++;159
z=1;160
c(x,y-1,z);161
}162
else if(a[x-1][y]=='D')163
{ printf("YES\n");164
p++;}165
else if(a[x-1][y]=='.')166
{167
a[x][y]='@';168
z=2;169
p++;170
c(x-1,y,z);171
}172
else if(a[x][y+1]=='D')173
{ printf("YES\n");174
p++;}175
else if(a[x][y+1]=='.')176
{177
a[x][y]='@';178
z=3;179
p++;180
c(x,y+1,z);181
}182
}183
if(p==0)184
printf("NO\n");185
}186
int main()187
{188
int T;189
int n,m;190
scanf("%d",&T);191
for(int i=1;i<=T;i++)192
{193
scanf("%d%d",&n,&m);194
for(int j=1;j<=n;j++){195
196
for(int k=1;k<=m;k++)197
{198
scanf("%c",&a[j][k]);199
200
} getchar();201
}202
for(int j=1;j<=n;j++)203
for(int k=1;k<=m;k++)204
{205
if(a[j][k]=='Y')206
{207
c(j,k,1);208
}209
}210
}211
return 0;212
}213
深搜的代码
1
#include<iostream>2
#include<stdio.h>3
using namespace std;4
char a[21][21];5
int aa[][4]={{0,1,2,3},{1,2,3,0},{2,3,0,1},{3,0,1,2}};6
int kk=0,flag=0;7
int x_end,y_end;8
void dfs(int x,int y)9
{10
11
int k=kk;12
if(x==x_end&&y==y_end)//不能用=='D' 因为在上一层已经将 D的值改变13
{14
15
flag=1;16
return ;17
}18
for(int i=0;i<4&&!flag;++i)19
{20
if(aa[k][i]==0&&a[x][y-1]!='#')21
{22
a[x][y-1]='#';23
kk=0;dfs(x,y-1);break;//不能回溯的 因为只有一条路可走24
}25
if(aa[k][i]==1&&a[x-1][y]!='#')26
{27
a[x-1][y]='#';kk=1;dfs(x-1,y);break;28
}29
if(aa[k][i]==2&&a[x][y+1]!='#')30
{31
a[x][y+1]='#';kk=2;dfs(x,y+1);break;32
}33
if(aa[k][i]==3&&a[x+1][y]!='#')34
{35
a[x+1][y]='#';kk=3;dfs(x+1,y);break;36
}37
}38
}39
int main()40
{41
//freopen("in.txt","r",stdin);42
int T;43
int n,m;44
scanf("%d",&T);45
for(int i=1;i<=T;i++)46
{47
kk=0;48
flag=0;//在循环的开始对kk和flag进行清零49
scanf("%d%d",&n,&m);50
getchar();51
for(int j=1;j<=n;j++)52
{53
for(int k=1;k<=m;k++)54
{55
scanf("%c",&a[j][k]);56
if(a[j][k]=='D')57
{58
x_end=j;59
y_end=k;60
}61
}62
getchar();63
}64
65
for(int j=1;j<=n;j++)66
for(int k=1;k<=m;k++)67
{68
if(a[j][k]=='Y')69
{70
a[j][k]='#';71
dfs(j,k);72
}73
}74
if(flag==1)75
printf("YES\n");76
else77
printf("NO\n");78
79
}80
return 0;81
}82