Colored Sticks--POJ 2513
1、解题思路：字典树，并查集，欧拉通路。
2、注意事项：字典树的插入；并查集与DFS的结合；欧拉回路的判断。
3、实现方法：
1
#include
<
iostream
>
2
#include
<
map
>
3
#include
<
string
>
4
using
namespace
std;
5
6
//
用邻接表表示连接关系
7
struct
Edge
8
{
9
int
v;
10
Edge
*
nxt;
11
};
12
Edge
*
head[
1000010
];
13
Edge Arr[
1000010
];
14
15
struct
Trie
16
{
17
bool
IsStr;
18
int
Id;
19
Trie
*
nxt[
26
];
20
};
21
Trie root,temp[
1000000
];
22
23
int
pos,cnt,num;
24
bool
visite[
500010
];
25
26
//
trie树的插入
27
int
Insert(
char
str[])
28
{
29
int
len
=
strlen(str);
30
Trie
*
Current
=&
root;
31
for
(
int
i
=
0
;i
<
len;i
++
)
32
{
33
if
(Current
->
nxt[str[i]
-
'
a
'
]
==
NULL)
34
Current
->
nxt[str[i]
-
'
a
'
]
=&
temp[pos
++
];
35
Current
=
Current
->
nxt[str[i]
-
'
a
'
];
36
}
37
if
(Current
->
IsStr)
38
return
Current
->
Id;
39
Current
->
IsStr
=
true
;
40
Current
->
Id
=++
cnt;
41
return
Current
->
Id;
42
}
43
44
void
DFS(
int
v)
45
{
46
Edge
*
w
=
head[v]
->
nxt;
47
visite[v]
=
true
;
48
while
(w)
49
{
50
if
(visite[w
->
v]
==
false
)
51
DFS(w
->
v);
52
w
=
w
->
nxt;
53
}
54
}
55
56
bool
Judge()
57
{
58
int
x
=
0
;
59
for
(
int
i
=
1
;i
<=
cnt;i
++
)
60
{
61
if
(visite[i]
==
false
)
62
{
63
DFS(i);
64
x
++
;
65
}
66
if
(x
>
1
)
67
return
false
;
68
}
69
return
true
;
70
}
71
72
void
Init()
73
{
74
char
str1[
11
],str2[
11
];
75
int
A1,A2;
76
while
(scanf(
"
%s %s
"
,str1,str2)
!=
EOF)
77
{
78
A1
=
Insert(str1);
79
A2
=
Insert(str2);
80
if
(head[A1]
==
NULL)
81
head[A1]
=&
Arr[num
++
];
82
if
(head[A2]
==
NULL)
83
head[A2]
=&
Arr[num
++
];
84
Edge
*
p
=
head[A1],
*
q;
85
while
(p)
86
{
87
q
=
p;p
=
p
->
nxt;
88
}
89
q
->
nxt
=&
Arr[num
++
];
90
q
->
nxt
->
v
=
A2;
91
Edge
*
r
=
head[A2],
*
w;
92
while
(r)
93
{
94
w
=
r;r
=
r
->
nxt;
95
}
96
w
->
nxt
=&
Arr[num
++
];
97
w
->
nxt
->
v
=
A1;
98
}
99
}
100
101
int
main()
102
{
103
int
i,s,ans
=
0
;
104
Init();
105
if
(cnt
==
0
)
106
cout
<<
"
Possible
"
<<
endl;
107
else
108
{
109
if
(Judge())
110
{
111
for
(i
=
1
;i
<=
cnt;i
++
)
112
{
113
Edge
*
p
=
head[i];
114
s
=-
1
;
115
while
(p)
116
{
117
s
++
; p
=
p
->
nxt;
118
}
119
if
(s
%
2
==
1
)
120
ans
++
;
121
}
122
if
(ans
==
0
||
ans
==
2
)
123
cout
<<
"
Possible
"
<<
endl;
124
else
125
cout
<<
"
Impossible
"
<<
endl;
126
}
127
else
128
cout
<<
"
Impossible
"
<<
endl;
129
}
130
return
0
;
131
}