TJU 1168. Team Queue
Time Limit: 1.0 Seconds Memory Limit: 65536K
Total Runs: 265 Accepted Runs: 108
本题总的来说不难，却困扰了我不少时间，因为我不知道原来数列竟然可以定义成数组队列。如果知道这个就很容易了。
思路：首先定义一个总的队列用来存储入队的那些TEAM的编号，然后定义一个数组队列，每一个队列用来记录本TEAM的入队顺序。首先是入队时判断所在TEAM的队列是否为空如若为空将TEAM的编号加入总队列中，然后再将本数加入到所在TEAM的队列中。出队时，判断所在TEAM的队列是否为空，如果为空总队列也要输出。
本题我本来是用链表做的，本以为链表会很快，但是后来发现题意是进队和出对都是常数时间，所以超时是理所当然的
链表代码：（超时）
代码
1
#include
<
stdio.h
>
2
#include
<
string
.h
>
3
#include
<
stdlib.h
>
4
struct
node
5
{
6
int
set
;
7
}s[
1000000
];
8
typedef
struct
t
9
{
10
int
set
;
11
int
data;
12
struct
t
*
next;
13
}
*
T;
14
int
main()
15
{
16
T head,p;
17
int
count
=
0
,t,i,j,n,a;
18
char
b[
10
];
19
while
(scanf(
"
%d
"
,
&
t)
!=
EOF)
20
{
21
count
++
;
22
if
(t
==
0
)
23
break
;
24
for
(i
=
1
;i
<=
t;i
++
)
25
{
26
scanf(
"
%d
"
,
&
n);
27
for
(j
=
1
;j
<=
n;j
++
)
28
{
29
scanf(
"
%d
"
,
&
a);
30
s[a].
set
=
i;
31
}
32
}
33
head
=
(T)malloc(
sizeof
(t));
34
head
->
next
=
NULL;
35
p
=
(T)malloc(
sizeof
(t));
36
//
p->next=NULL;
37
printf(
"
Scenario #%d\n
"
,count);
38
while
(scanf(
"
%s
"
,b)
!=
EOF)
39
{
40
if
(strcmp(b,
"
STOP
"
)
==
0
)
41
break
;
42
else
if
(b[
0
]
==
'
E
'
)
43
{
44
scanf(
"
%d
"
,
&
a);
45
T t
=
(T)malloc(
sizeof
(t));
46
t
->
next
=
NULL;
47
t
->
set
=
s[a].
set
;
48
t
->
data
=
a;
49
p
=
head;
50
while
(p
->
next
!=
NULL)
51
{
52
if
(p
->
set
==
s[a].
set
&&
p
->
next
->
set
!=
s[a].
set
)
53
break
;
54
p
=
p
->
next;
55
}
56
t
->
next
=
p
->
next;
57
p
->
next
=
t;
58
}
59
else
if
(b[
0
]
==
'
D
'
)
60
{
61
p
=
head
->
next;
62
printf(
"
%d\n
"
,p
->
data);
63
head
->
next
=
head
->
next
->
next;
64
}
65
}
66
67
}
68
return
0
;
69
}
代码：
1
#include
<
stdio.h
>
2
#include
<
queue
>
3
using
namespace
std;
4
struct
node
5
{
6
int
data;
7
int
set
;
8
}s[
1000000
];
9
int
main()
10
{
11
int
count
=
0
,t,i,j,a,n;
12
char
b[
10
];node cur1;
13
while
(scanf(
"
%d
"
,
&
t)
!=
EOF)
14
{
15
if
(t
==
0
)
16
break
;
17
count
++
;
18
int
cur;
19
for
(i
=
1
;i
<=
t;i
++
)
20
{
21
scanf(
"
%d
"
,
&
n);
22
for
(j
=
1
;j
<=
n;j
++
)
23
{
24
scanf(
"
%d
"
,
&
a);
25
s[a].
set
=
i;
26
s[a].data
=
a;
27
}
28
}
29
printf(
"
Scenario #%d\n
"
,count);
30
queue
<
int
>
q;
31
queue
<
node
>
qu[
1000
];
32
while
(scanf(
"
%s
"
,b)
!=
EOF)
33
{
34
if
(strcmp(b,
"
STOP
"
)
==
0
)
35
break
;
36
else
if
(b[
0
]
==
'
E
'
)
37
{
38
scanf(
"
%d
"
,
&
a);
39
if
(qu[s[a].
set
].empty ())
40
q.push(s[a].
set
);
41
qu[s[a].
set
].push(s[a]);
42
}
43
else
if
(b[
0
]
==
'
D
'
)
44
{
45
cur
=
q.front ();
46
cur1
=
qu[cur].front();
47
qu[cur].pop();
48
if
(qu[cur].empty ())
49
q.pop();
50
printf(
"
%d\n
"
,cur1.data);
51
}
52
}
53
printf(
"
\n
"
);
54
}
55
return
0
;
56
}
57