| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5303 | Accepted: 3162 |
这一题相当得简单只要正反比较两次即可，运用队列表示正向的，用栈表示反向的
代码：
1
#include
<
stdio.h
>
2
#include
<
queue
>
3
#include
<
stack
>
4
using
namespace
std;
5
int
main()
6
{
7
int
t,i,mark;
8
char
s1[
105
],s2[
105
];
9
10
scanf(
"
%d
"
,
&
t);
11
while
(t
--
)
12
{
13
mark
=
0
;
14
scanf(
"
%s%s
"
,s1,s2);
15
queue
<
char
>
qu;
16
stack
<
char
>
st;
17
for
(i
=
0
;s2[i]
!=
'
\0
'
;i
++
)
18
{
19
st.push(s2[i]);
20
qu.push(s2[i]);
21
}
22
for
(i
=
0
;s1[i]
!=
'
\0
'
;i
++
)
23
{
24
char
x,y;
25
x
=
st.top();
26
y
=
qu.front();
27
if
(x
==
s1[i])
28
st.pop();
29
if
(st.empty())
30
{
31
mark
=
1
;
32
break
;
33
}
34
if
(y
==
s1[i])
35
qu.pop();
36
if
(qu.empty())
37
{
38
mark
=
1
;
39
break
;
40
}
41
}
42
if
(mark)
43
printf(
"
YES\n
"
);
44
else
45
printf(
"
NO\n
"
);
46
}
47
return
0
;
48
}
49
