POJ 3278 Catch That Cow
解题思路:BFS
代码
#include
<
iostream
>
#include
<
cmath
>
using
namespace
std;
#define
MAXN 100001
int
n, k, q[MAXN];
bool
visit[MAXN];
int
main()
{
int
s, e, p, i, j,step,minStep;
while
(scanf(
"
%d %d
"
,
&
n,
&
k)
!=
EOF)
{
q[
0
]
=
n,s
=
e
=
step
=
0
,minStep
=
MAXN;
if
(n
>=
k)minStep
=
step
=
n
-
k;
memset(visit,
0
,
sizeof
(visit));
while
(s
<=
e
&&
step
<
minStep)
{
for
(i
=
s,p
=
e;i
<=
e;i
++
)
{
if
(q[i]
==
k
&&
minStep
>
step)minStep
=
step;
if
(q[i]
-
1
>=
0
&&!
visit[q[i]
-
1
])q[
++
p]
=
q[i]
-
1
,visit[q[p]]
=
true
;
if
(q[i]
+
1
<=
k
&&!
visit[q[i]
+
1
])q[
++
p]
=
q[i]
+
1
, visit[q[p]]
=
true
;
if
(
2
*
q[i]
<=
k
&&!
visit[
2
*
q[i]])q[
++
p]
=
2
*
q[i],visit[q[p]]
=
true
;
else
if
(
2
*
q[i]
>
k)
{
if
(
2
*
q[i]
-
k
+
step
+
1
<
minStep)minStep
=
2
*
q[i]
-
k
+
step
+
1
;
if
(k
-
q[i]
+
step
<
minStep)minStep
=
k
-
q[i]
+
step;
}
}
step
++
,s
=
e
+
1
,e
=
p;
}
printf(
"
%d\n
"
, minStep);
}
return
0
;
}