The Accomodation of Students
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1462 Accepted Submission(s): 716
Problem Description
There are a group of students. Some of them may know each other, while others don't. For example, A and B know each other, B and C know each other. But this may not imply that A and C know each other.
Now you are given all pairs of students who know each other. Your task is to divide the students into two groups so that any two students in the same group don't know each other.If this goal can be achieved, then arrange them into double rooms. Remember, only paris appearing in the previous given set can live in the same room, which means only known students can live in the same room.
Calculate the maximum number of pairs that can be arranged into these double rooms.
Input
For each data set:
The first line gives two integers, n and m(1<n<=200), indicating there are n students and m pairs of students who know each other. The next m lines give such pairs.
Proceed to the end of file.
Output
If these students cannot be divided into two groups, print "No". Otherwise, print the maximum number of pairs that can be arranged in those rooms.
Sample Input
4 4
1 2
1 3
1 4
2 3
6 5
1 2
1 3
1 4
2 5
3 6
Sample Output
No
3
Source
2008 Asia Harbin Regional Contest Online
Recommend
gaojie
又是一遍AC,虽然还是模板题,但是一个小时4个1a让我今天下午很愉快啊有木有.
#include<stdio.h>
#include
<
string
.h>
int
N,M;
int
color[
300
],match[
300
];
bool
visit[
300
],G[
300
][
300
],flag;
void
draw(
int
k,
int
cc) {
if
(color[k]!=-
1
&& color[k]!=
cc) { flag
=
false
;
return
; }
if
(color[k]==cc)
return
; color[k]
=
cc;
int
c=
1
-
cc;
for
(
int
i=
1
;i<=N;i++
)
if
(G[k][i] &&
flag) draw(i,c); }
bool
DFS(
int
k) {
int
t;
for
(
int
i=
1
;i<=N;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=
1
;i<=N;i++
) { memset(visit,
0
,
sizeof
(visit));
if
(DFS(i)) ans++
; }
return
ans; }
int
main() {
while
(scanf(
"
%d%d
"
,&N,&M)!=
EOF) { memset(G,
0
,
sizeof
(G));
for
(
int
i=
1
;i<=M;i++
) {
int
u,v; scanf(
"
%d%d
"
,&u,&
v); G[u][v]
=
1
; G[v][u]
=
1
; } memset(color,
-
1
,
sizeof
(color)); flag
=
true
;
for
(
int
i=
1
;i<=N;i++
)
if
(flag && color[i]==-
1
) draw(i,
0
);
if
(!flag) printf(
"
No\n
"
);
else
printf(
"
%d\n
"
,Max_match()/
2
); }
return
0
; }
