uva 442 - Matrix Chain Multiplication
表示从今以后要好好学习，代码写的太烂了
题目大意是求最少的计算次数，唉。。。
最大收获是学会了求表达式中各部分的优先级
#include
<
stdio.h
>
#include
<
stdlib.h
>
typedef
struct
_matrix
{
char
c;
int
row;
int
col;
int
rank;
struct
_matrix
*
next;
struct
_matrix
*
last;
}matrix;
matrix symbol[
30
];
char
s[
200
];
char
*
_s;
int
i,n,rank,first,error,sum,max;
matrix
*
head,
*
now,
*
p,
*
_p,
*
pp,
*
end;
void
getsum()
{
error
=
0
;
for
(i
=
max;i
>=
0
&&
head
->
next
!=
NULL;i
--
)
{
p
=
head;
while
(p
->
rank
!=
i
&&
p
!=
NULL)
p
=
p
->
next;
if
(p
->
next
==
NULL
||
p
->
rank
!=
p
->
next
->
rank)
{
p
->
rank
--
;
continue
;
}
while
(p
->
next
!=
NULL
&&
p
->
rank
==
p
->
next
->
rank)
{
if
(p
->
col
!=
p
->
next
->
row)
{
error
=
1
;
break
;
}
sum
+=
(p
->
row)
*
(p
->
col)
*
(p
->
next
->
col);
now
=
(matrix
*
)malloc(
sizeof
(matrix));
now
->
row
=
p
->
row;
now
->
col
=
p
->
next
->
col;
now
->
rank
=
p
->
rank;
_p
=
p
->
last;
pp
=
p
->
next
->
next;
if
(head
->
next
->
next
==
NULL)
{
free(p
->
next);
free(p);
head
=
now;
head
->
next
=
NULL;
p
=
head;
break
;
}
else
if
(p
==
head)
{
head
=
now;
now
->
next
=
pp;
pp
->
last
=
now;
head
->
last
=
NULL;
}
else
if
(p
->
next
==
end)
{
end
=
now;
end
->
next
=
NULL;
_p
->
next
=
now;
now
->
last
=
_p;
}
else
{
_p
->
next
=
now;
now
->
next
=
pp;
pp
->
last
=
now;
now
->
last
=
_p;
}
free(p
->
next);
free(p);
p
=
now;
}
now
->
rank
--
;
if
(error)
break
;
}
}
int
main()
{
scanf(
"
%d
"
,
&
n);
for
(i
=
0
;i
<
n;i
++
)
{
getchar();
scanf(
"
%c%d%d
"
,
&
symbol[i].c,
&
symbol[i].row,
&
symbol[i].col);
}
getchar();
while
(gets(s))
{
_s
=
s;
rank
=
0
;
first
=
1
;
max
=
0
;
sum
=
0
;
while
(
*
_s)
{
if
(
*
_s
==
'
(
'
)
{
rank
++
;
if
(rank
>
max)
max
=
rank;
}
else
if
(
*
_s
==
'
)
'
)
{
rank
--
;
if
(
*
(_s
+
1
)
==
'
(
'
)
{
getsum();
if
(error)
break
;
head
->
rank
=
0
;
rank
=
0
;
max
=
0
;
first
=
2
;
}
}
else
{
now
=
(matrix
*
)malloc(
sizeof
(matrix));
for
(i
=
0
;
*
_s
!=
symbol[i].c;i
++
);
now
->
c
=*
_s;
now
->
rank
=
rank;
now
->
row
=
symbol[i].row;
now
->
col
=
symbol[i].col;
if
(first
==
1
)
{
head
=
now;
head
->
next
=
NULL;
head
->
last
=
NULL;
p
=
head;
end
=
p;
first
=
0
;
}
else
{
now
->
last
=
p;
p
->
next
=
now;
p
=
now;
end
=
p;
p
->
next
=
NULL;
}
}
_s
++
;
}
getsum();
if
(error)
printf(
"
error\n
"
);
else
printf(
"
%d\n
"
,sum);
}
return
0
;
}