POJ 3126 Prime Path
解题思路:
1）首先计算0-9999之间的素数
2）BFS，每次更改4位数字中的一位数字(首位数字不能为0)，判断是否为素数，是且未被访问过的加入到队列里面
3）队列为空，或者发现目的素数，停止
NULL
#include
<
iostream
>
using
namespace
std;
#define
MAXN 10000
int
main()
{
int
i,j,k,l,n,p,q,s,e,iter,ts,num,radix,qu[MAXN],d[
4
];
bool
IsPrime[MAXN],visit[MAXN];
memset(IsPrime,
1
,
sizeof
(IsPrime));
IsPrime[
0
]
=
IsPrime[
1
]
=
false
;
for
(i
=
2
;i
<
5001
;i
++
)
if
(IsPrime[i])
for
(j
=
i
*
2
;j
<
MAXN;j
+=
i)IsPrime[j]
=
false
;
scanf(
"
%d
"
,
&
n);
while
(n
--
)
{
scanf(
"
%d %d
"
,
&
p,
&
q);
memset(visit,
0
,
sizeof
(visit));
qu[
0
]
=
p,iter
=
s
=
e
=
0
,visit[p]
=
true
;
while
(s
<=
e
&&
!
visit[q])
{
for
(ts
=
e,i
=
s;i
<=
e
&&!
visit[q];i
++
)
{
for
(num
=
qu[i],j
=
3
,radix
=
1000
;j
>=
0
;j
--
)d[j]
=
num
/
radix,num
-=
d[j]
*
radix,radix
/=
10
;
for
(j
=
0
;j
<
4
;j
++
)
for
(k
=
0
;k
<
10
;k
++
)
{
if
(j
==
3
&&
k
==
0
)
continue
;
for
(radix
=
1
,num
=
l
=
0
;l
<
4
;radix
*=
10
,l
++
)
if
(l
==
j)num
+=
k
*
radix;
else
num
+=
d[l]
*
radix;
if
(
!
visit[num]
&&
IsPrime[num])qu[
++
ts]
=
num, visit[num]
=
true
;
}
}
s
=
e
+
1
,e
=
ts,iter
++
;
}
visit[q]
?
printf(
"
%d\n
"
, iter):printf(
"
Impossible\n
"
);
}
return
0
;
}