Prime Path
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 71 Accepted Submission(s) : 32
Problem Description
The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices.
— It is a matter of security to change such things every now and then, to keep the enemy in the dark.
— But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know!
— I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door.
— No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime!
— I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds.
— Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime.
Now, the minister of finance, who had been eavesdropping, intervened.
— No unnecessary expenditure, please! I happen to know that the price of a digit is one pound.
— Hmm, in that case I need a computer program to minimize the cost. You don't know some very cheap software gurus, do you?
— In fact, I do. You see, there is this programming contest going on... Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above.
1033
1733
3733
3739
3779
8779
8179
The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased.
Input
One line with a positive number: the number of test cases (at most 100). Then for each test case, one line with two numbers separated by a blank. Both numbers are four-digit primes (without leading zeros).
Output
One line for each case, either with a number stating the minimal cost or containing the word Impossible.
Sample Input
3
1033 8179
1373 8017
1033 1033
Sample Output
6
7
0
题目：给两个四位数 （都是素数），每一次只能变一个数字，变换后也得是素数 问第一个数字几下可以变成第二个素数 不能则输出impossible
思路：
bfs 把每个位置每个数字都加入队列，暴力测试，注意千位的变化 ，0 的时候要想到，没注意这点就得wa到死
当然要先打个素数表，如果判断的话容易超时，
Source Code#include<iostream>
#include<stdio.h>
#include<string.h>
#include<queue>
#include<string>
using namespace std
;
void change(int c
,char m
[])
{
m
[0]=c
/1000;
m
[1]=c
%1000/100;
m
[2]=c
%100/10;
m
[3]=c
%10;
}
int vis
[10001];
int a
,b
;
int ans
;
int judge(int n
)
{
int i
;
for(i
=2;i
<=n
/2;i
++)
{
if(n
%i
==0)
{
return 0;
}
}
return 1;
}
int p
[10001];
void findprime()
{
for(int i
=2;i
<10001;i
++)
{
if(!p
[i
])
{
for(int j
=2;i
*j
<10001;j
++)
p
[i
*j
]=1;
}
}
}
struct node
{
int x
;
int step
;
};
int bfs()
{
int i
,j
;
char m
[4];
queue
<node
> Q
;
node now
,next
;
now
.x
=a
;
now
.step
=0;
Q
.push(now
);
vis
[a
]=1;
while(!Q
.empty())
{
now
=Q
.front();
Q
.pop();
if(now
.x
==b
)
{
ans
=now
.step
;
return 0;
}
for(i
=1;i
<=4;i
++)
{
change(now
.x
,m
);
for(j
=0;j
<10;j
++)
{
if(j
==0&&i
==1)continue;
if(i
==1)next
.x
=j
*1000+m
[1]*100+m
[2]*10+m
[3];
if(i
==2)next
.x
=m
[0]*1000+j
*100+m
[2]*10+m
[3];
if(i
==3)next
.x
=m
[0]*1000+m
[1]*100+j
*10+m
[3];
if(i
==4)next
.x
=m
[0]*1000+m
[1]*100+m
[2]*10+j
;
if(!p
[next
.x
]&&!vis
[next
.x
])
{
next
.step
=now
.step
+1;
vis
[next
.x
]=1;
Q
.push(next
);
}
}
}
}
return 0;
}
int main()
{
int i
,j
;
int t
;
cin
>>t
;
p
[1]=0;
p
[2]=0;
p
[3]=0;
findprime();
while(t
--)
{
ans
=0;
memset(vis
,0,sizeof(vis
));
cin
>>a
>>b
;
bfs();
if(a
==b
)
cout
<<0<<endl
;
else if(ans
!=0)
cout
<<ans
<<endl
;
else cout
<<"Impossible"<<endl
;
}
return 0;
}
