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
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.
The input contains several data sets in text format. Each data set represents one set of subjects of the study, with the following description:
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)
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.
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
Sample Output
5
2
Source
Southeastern Europe 2000
Recommend
JGShining
嗯,看起来是个求最大独立集的题,而且还是个二分图.因为题里没说有 Homosexuality,所以不会有奇环.不过图是用一般形式给出的算出来的匹配要除以2.
#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
);
}
}
