hdu 1053 Entropy
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 802 Accepted Submission(s): 299
这是作者第一次写赫夫曼，虽然WRONG了几次，但是最终把它做出来了，也算是一种小小的成就吧。本题的思想就是根据所给的字符串判断每个字符的权重，然后再根据每个字符的权重来构建赫夫曼树，最后对赫夫曼树遍历，判断每个字符所在的层数，所有字符的层数*个数就等于压缩后的大小：
赫夫曼的构建思想是：将所有的节点放入到优先队列里面，对优先队列进行重载，是从小到大排列。每次取出优先队列前两个数。组成一个数之后再放进优先队列。依次循环下去，直到只剩下一个的时候便是根节点。
代码：
1
#include
<
stdio.h
>
2
#include
<
queue
>
3
#include
<
string
.h
>
4
using
namespace
std;
5
int
k[
1000
];
6
typedef
struct
T
7
{
8
char
data; // 记录字符
9
bool
o; // 记录是字符还是由字符重组的
10
struct
T
*
Lchild,
*
Rchild;
11
}
*
Tree;
12
typedef
struct
node
13
{
14
int
n; // 记录每个字符的权重
15
int
h; // 记录是字符还是由字符重组的
16
Tree t;
17
bool
operator
<
(
const
node
&
a)
const
18
{ //对优先队列进行重载，权重从小到大如果权重相同，重组的让给单个
19
if
(a.n
!=
n)
20
return
a.n
<
n;
21
else
22
return
a.h
<
h;
23
}
24
}NODE;
25
26
NODE cur1,cur2,cur;
27
priority_queue
<
NODE
>
qu;
28
void
createTree()
29
{
30
cur1
=
qu.top();
31
qu.pop();
32
if
(qu.empty())
33
return
;
34
cur2
=
qu.top();
35
qu.pop();
36
cur.n
=
cur1.n
+
cur2.n;
37
cur.h
=
1
;
38
cur.t
=
(Tree)malloc(
sizeof
(T));
39
cur.t
->
Lchild
=
cur1.t;
40
cur.t
->
Rchild
=
cur2.t;
41
cur.t
->
data
=
'
'
;
42
cur.t
->
o
=
false
;
43
qu.push(cur);
44
createTree();
45
}
46
void
visit(Tree c,
int
floor)
47
{
48
if
(c
->
o
==
true
)
49
{
50
if
(c
->
data
==
'
_
'
)
51
k[
0
]
=
floor;
52
else
53
k[c
->
data
-
'
A
'
+
1
]
=
floor;
54
}
55
if
(c
->
Lchild
!=
NULL)
56
visit(c
->
Lchild,floor
+
1
);
57
if
(c
->
Rchild
!=
NULL)
58
visit(c
->
Rchild,floor
+
1
);
59
}
60
int
main()
61
{
62
char
a[
1000
];
63
int
i,j,mark[
1000
],len,count,sum;
64
while
(scanf(
"
%s
"
,a)
!=
EOF)
65
{
66
sum
=
0
;
67
while
(
!
qu.empty())
68
qu.pop();
69
memset(mark,
0
,
sizeof
(mark));
70
memset(k,
0
,
sizeof
(k));
71
if
(strcmp(a,
"
END
"
)
==
0
)
72
break
;
73
len
=
strlen(a);
74
for
(i
=
0
;i
<
len;i
++
)
75
{
76
count
=
0
;
77
if
(
!
mark[i])
78
{
79
for
(j
=
i;j
<
len;j
++
)
80
{
81
if
(a[i]
==
a[j])
82
{
83
mark[j]
=
1
;
84
count
++
;
85
}
86
}
87
cur1.n
=
count;
88
cur1.h
=
0
;
89
cur1.t
=
(Tree)malloc(
sizeof
(T));
90
cur1.t
->
data
=
a[i];
91
cur1.t
->
Lchild
=
NULL;
92
cur1.t
->
Rchild
=
NULL;
93
cur1.t
->
o
=
true
;
94
qu.push(cur1);
95
}
96
}
97
if
(qu.size()
==
1
)
98
{
99
printf(
"
%d %d 8.0\n
"
,len
*
8
,len);
100
continue
;
101
}
102
createTree();
103
visit(cur1.t,
0
);
104
for
(i
=
0
;i
<
len;i
++
)
105
if
(a[i]
==
'
_
'
)
106
sum
+=
k[
0
];
107
else
108
sum
+=
k[a[i]
-
'
A
'
+
1
];
109
printf(
"
%d %d %.1lf\n
"
,len
*
8
,sum,len
*
8
*
1.0
/
sum);
110
}
111
return
0
;
112
}
113
114
115