POJ 2912 Rochambeau
解题思路:并查集
并查集的思想与POJ1182 食物链相同
枚举所有Judge ID 情况：
如果有且仅有一个ID没有产生矛盾，则说明该ID即为Judge，且发现的最初回合为其余首次发现矛盾回合的最大值
如果没有ID产生矛盾，输出Can not determine
如果有一个以上ID产生矛盾，输出Impossible
欢迎指教
1
#include
<
iostream
>
2
using
namespace
std;
3
4
#define
MAXN 501
5
#define
MAXM 2001
6
int
root[MAXN],relation[MAXN],x[MAXM],y[MAXM],sa[MAXN];
7
char
cmp[MAXM];
8
int
findroot(
int
x)
9
{
10
int
t;
11
if
(root[x]
!=
x)
12
{
13
t
=
root[x];root[x]
=
findroot(root[x]);
14
relation[x]
=
(relation[x]
+
relation[t])
%
3
;
15
}
16
return
root[x];
17
}
18
19
bool
IsTrue(
int
x,
int
y,
int
d)
20
{
21
int
a, b;
22
a
=
findroot(x);
23
b
=
findroot(y);
24
if
(a
==
b)
25
return
(
3
+
relation[x]
-
relation[y])
%
3
==
d
?
true
:
false
;
26
else
27
root[b]
=
a,relation[b]
=
(
6
-
d
+
relation[x]
-
relation[y])
%
3
;
28
return
true
;
29
}
30
31
int
main()
32
{
33
int
i,N,M,judge,r,t,ansP,ansS;
34
while
(scanf(
"
%d %d
"
,
&
N,
&
M)
==
2
)
35
{
36
memset(sa,
0
,
sizeof
(sa));
37
for
(i
=
0
;i
<
M;i
++
)scanf(
"
%d %c %d
"
,
&
x[i],
&
cmp[i],
&
y[i]);
38
for
(judge
=
0
;judge
<
N;judge
++
)
39
{
40
for
(i
=
0
;i
<
N;i
++
)root[i]
=
i,relation[i]
=
0
;
41
for
(i
=
0
;i
<
M;i
++
)
42
{
43
if
(x[i]
==
judge
||
y[i]
==
judge)
continue
;
44
if
(cmp[i]
==
'
=
'
)r
=
0
;
45
else
if
(cmp[i]
==
'
>
'
)r
=
1
;
46
else
r
=
2
;
47
if
(
!
IsTrue(x[i],y[i],r)){sa[judge]
=
i
+
1
;
break
;}
48
}
49
}
50
for
(ansS
=
i
=
t
=
0
;i
<
N;i
++
)
51
{
52
if
(sa[i]
==
0
)t
++
,ansP
=
i;
53
if
(sa[i]
>
ansS)ansS
=
sa[i];
54
}
55
if
(t
==
0
)printf(
"
Impossible\n
"
);
56
else
if
(t
>
1
)printf(
"
Can not determine\n
"
);
57
else
printf(
"
Player %d can be determined to be the judge after %d lines\n
"
,ansP,ansS);
58
}
59
return
0
;
60
}
61