Blue Jeans--POJ 3080
1、题目类型：字符串、暴力法、KMP算法。
2、解题思路：寻找最长的公共子串，STL中string部分方法的应用。
3、注意事项：string.h中库函数的调用，KMP匹配。
4、实现方法： （暴力法）
1
#include
<
iostream
>
2
#include
<
string
>
3
using
namespace
std;
4
5
string
str[
10
];
6
7
int
check(
int
nStr,
string
s)
8
{
9
int
i;
10
for
(i
=
1
; i
<
nStr; i
++
)
11
{
12
//
string的find()函数当没有找到相应子串时返回sting::npos
13
if
(str[i].find(s)
==
string
::npos)
14
return
0
;
15
}
16
return
1
;
17
}
18
19
string
gene(
int
nStr)
20
{
21
int
len, j;
22
string
longest;
23
for
(len
=
str[
0
].length(); len
>=
3
; len
--
)
24
{
25
for
(j
=
0
; j
<=
str[
0
].length()
-
len; j
++
)
26
{
27
string
substring
=
str[
0
].substr(j, len);
//
从j开始获得len长度的字符串
28
if
(check(nStr, substring))
29
//
对比新旧获得子串
30
if
(substring.size()
>
longest.size()
||
(substring.size()
==
longest.size()
&&
substring
<
longest))
31
longest
=
substring;
32
}
33
}
34
return
longest;
35
}
36
37
void
Seach(
int
nStr)
38
{
39
string
substring
=
gene(nStr);
40
if
(substring.size())
41
cout
<<
substring
<<
endl;
42
else
cout
<<
"
no significant commonalities
"
<<
endl;
43
}
44
int
main()
45
{
46
int
nCase;
47
int
nStr, i;
48
cin
>>
nCase;
49
while
(nCase
--
)
50
{
51
cin
>>
nStr;
52
for
(i
=
0
; i
<
nStr; i
++
)
53
cin
>>
str[i];
54
Seach(nStr);
55
}
56
return
0
;
57
}
58
4、实现方法： （KMP）
1
#include
<
iostream
>
2
#include
<
string
>
3
using
namespace
std;
4
5
const
int
N
=
70
,M
=
15
;
6
int
ls,lp,pre[N],len;
7
char
str[M][N],s2[N],ans[N];
8
9
void
prefix(
char
*
p)
10
{
11
int
i, k;
12
memset(pre,
0
,
sizeof
(pre));
13
pre[
1
]
=
0
;
14
k
=
0
;
15
for
(i
=
2
;i
<=
lp;i
++
)
16
{
17
while
(k
>
0
&&
p[k
+
1
]
!=
p[i])
18
k
=
pre[k];
19
if
(p[k
+
1
]
==
p[i]) k
++
;
20
pre[i]
=
k;
21
}
22
}
23
24
bool
kmp(
char
*
p,
char
*
s)
25
{
26
int
i,k,cnt
=
0
;
27
k
=
0
;
28
for
(i
=
1
;i
<=
ls;i
++
)
29
{
30
while
(k
>
0
&&
p[k
+
1
]
!=
s[i])
31
k
=
pre[k];
32
if
(p[k
+
1
]
==
s[i])
33
k
++
;
34
if
(k
==
lp)
35
return
true
;
36
}
37
return
false
;
38
}
39
40
int
main()
41
{
42
int
casenum, strnum, i, j;
43
cin
>>
casenum;
44
while
(casenum
--
)
45
{
46
cin
>>
strnum;
47
for
(i
=
0
; i
<
strnum; i
++
)
48
cin
>>
str[i]
+
1
;
49
len
=
0
;
50
strcpy(ans,
"
no significant commonalities
"
);
51
for
(lp
=
60
;lp
>=
3
;lp
--
)
52
{
53
for
(i
=
1
;i
+
lp
-
1
<=
60
;i
++
)
54
{
55
prefix(str[
0
]
+
i
-
1
);
56
ls
=
60
;
57
for
(j
=
1
;j
<
strnum;j
++
)
58
{
59
if
(
!
kmp(str[
0
]
+
i
-
1
,str[j]))
60
break
;
61
}
62
if
(j
>=
strnum
&&
lp
>=
len)
63
{
64
strncpy(s2,str[
0
]
+
i,lp);
65
s2[lp]
=
0
;
66
if
(lp
>
len
||
lp
==
len
&&
strcmp(s2,ans)
<
0
)
67
strcpy(ans, s2);
68
}
69
}
70
if
(strcmp(ans,
"
no significant commonalities
"
)
!=
0
)
71
break
;
72
}
73
puts(ans);
74
}
75
return
0
;
76
}