Girls and Boys
Time Limit: 20000/10000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5565????Accepted Submission(s): 2481
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Problem Description
the second year of the university somebody started a study on the romantic relations between the students. The relation “romantically involved” is defined between one girl and one boy. For the study reasons it is necessary to find out the maximum set satisfying the condition: there are no two students in the set who have been “romantically involved”. The result of the program is the number of students in such a set.
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The input contains several data sets in text format. Each data set represents one set of subjects of the study, with the following description:
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the number of students
the description of each student, in the following format
student_identifier:(number_of_romantic_relations) student_identifier1 student_identifier2 student_identifier3 ...
or
student_identifier:(0)
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The student_identifier is an integer number between 0 and n-1, for n subjects.
For each given data set, the program should write to standard output a line containing the result.
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Sample Input
7
0: (3) 4 5 6
1: (2) 4 6
2: (0)
3: (0)
4: (2) 0 1
5: (1) 0
6: (2) 0 1
3
0: (2) 1 2
1: (1) 0
2: (1) 0
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Sample Output
5
2
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Source
Southeastern Europe 2000
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Recommend
JGShining
嗯,看起來是個求最大獨立集的題,而且還是個二分圖.因為題里沒說有 Homosexuality,所以不會有奇環(huán).不過圖是用一般形式給出的算出來的匹配要除以2.
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#include<stdio.h>
#include
<
string
.h>
int
n,m,match[
1024
];
bool
visit[
1024
],G[
1024
][
1024
];
bool
DFS(
int
k)
{
int
t;
for
(
int
i=
0
;i<m;i++
)
if
(G[k][i] && !
visit[i])
{
visit[i]
=
1
;
t
=
match[i];
match[i]
=
k;
if
(t==-
1
|| DFS(t))
return
true
;
match[i]
=
t;
}
return
false
;
}
int
Max_match()
{
int
ans=
0
;
memset(match,
-
1
,
sizeof
(match));
for
(
int
i=
0
;i<n;i++
)
{
memset(visit,
0
,
sizeof
(visit));
if
(DFS(i)) ans++
;
}
return
ans;
}
int
main()
{
int
t,a,b,c;
while
(scanf(
"
%d
"
,&t)!=
EOF)
{
n
=m=
t;
memset(G,
0
,
sizeof
(G));
while
(t--
)
{
scanf(
"
%d: (%d)
"
,&a,&
b);
for
(
int
i=
0
;i<b;i++
)
{
scanf(
"
%d
"
,&
c);
G[a][c]
=
1
;
}
}
printf(
"
%d\n
"
,n-Max_match()/
2
);
}
}
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