PKU 2239 Selecting Courses(二分图最大匹配)
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题意大概是学校有多门课。没门课可能会在一个星期的每一天i(
1
<=
i
<=
7
)的第j(
1
<=
j
<=
12
)节课进行。求一个选课方案使得能选的课最多。
构造二分图,M
=
{课程}，N
=
{课程举行的时间},然后构造M到N得二分图。这里将时间(i,j)映射为一个整数(i
-
1
)
*
12
+
j用于构图。然后用匈牙利算法进行匹配就行了。
#include
<
iostream
>
#include
<
cstring
>
using
namespace
std;
int
map[
305
][
86
],match[
86
],course[
10
][
15
];
bool
vis[
86
];
int
n,t;
bool
findPath(
int
k)
{
for
(
int
i
=
1
;i
<=
84
;
++
i)
{
if
(
!
vis[i]
&&
map[k][i])
{
vis[i]
=
true
;
if
(
!
match[i]
||
findPath(match[i]))
{
match[i]
=
k;
return
true
;
}
}
}
return
false
;
}
int
main()
{
int
i,j,time1,time2;
for
(i
=
1
;i
<=
7
;
++
i)
{
for
(j
=
1
;j
<=
12
;
++
j) course[i][j]
=
(i
-
1
)
*
12
+
j;
}
while
(scanf(
"
%d
"
,
&
n)
!=
EOF)
{
memset(map,
0
,
sizeof
(map));
for
(i
=
1
;i
<=
n;
++
i)
{
scanf(
"
%d
"
,
&
t);
while
(t
--
)
{
scanf(
"
%d %d
"
,
&
time1,
&
time2);
map[i][course[time1][time2]]
=
1
;
}
}
int
ans
=
0
; memset(match,
0
,
sizeof
(match));
for
(i
=
1
;i
<=
n;
++
i)
{
memset(vis,
false
,
sizeof
(vis));
if
(findPath(i)) ans
++
;
}
printf(
"
%d\n
"
,ans);
}
return
0
;
}