Prime Path
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 27132   Accepted: 14861

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

Source

 
 
题目意思:
给出两个四位数的素数,要求对第一个素数进行一系列操作,使他变成第二个四位数的素数,每次操作只能改变一位数,且要求每次操作之后的数仍然是素数
问你最少的操作步骤数是多少?
 
分析:
先对四位数的素数打一个表
然后bfs
需要注意的是每次搜索入队之后,要进行回退操作
注意设置标记位(该数已经搜索过)
 
code:
#include<stdio.h>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <math.h>
#include <cstdlib>
#include <queue>
using namespace std;
#define max_v 10100
int prime[max_v];
int vis[max_v];
struct node
{
int x,step;
};
void init()//打N以内的素数表,prime[i]=0为素数。
{
memset(prime,,sizeof(prime));
prime[]=;
for(int i=; i<max_v; i++)
{
if(prime[i]==)
{
for(int j=; j*i<max_v; j++)
prime[j*i]=;
}
}
}
int bfs(int s,int e)
{
int num;
memset(vis,,sizeof(vis));//vis[N]用来标记是否查找过
queue<node> q; node p,next; p.x=s;
p.step=; vis[s]=;
q.push(p); while(!q.empty())
{
p=q.front();
q.pop(); if(p.x==e)
{
return p.step;
}
int t[];
t[]=p.x/;//千
t[]=p.x/%;//百
t[]=p.x/%;//十
t[]=p.x%;//个 for(int i=; i<=; i++)
{
int temp=t[i];//这里特别注意,每次只能变换一位数,这里用来保存该变换位的数,变换下一位数时,该位数要恢复
for(int j=; j<; j++)
{
if(t[i]!=j)
{
t[i]=j;//换数
num=t[]*+t[]*+t[]*+t[];
}
if(num>=&&num<=&&vis[num]==&&prime[num]==)//满足要求,四位数,没有用过,是素数
{
next.x=num;
next.step=p.step+;
q.push(next);
vis[num]=;
}
}
t[i]=temp;//恢复
}
}
return -;
}
int main()
{
//题意:给你两个4位数,要求你每次只能变换一位数字,并且变换前后的数字都要满足是素数;求最少变换次数;
//首先打好素数表,再进行BFS
int n,a,b,ans;
init();
cin>>n;
while(n--)
{
cin>>a>>b;
ans=bfs(a,b);
if(ans==-)
{
cout<<"Impossible"<<endl;
}
else
{
cout<<ans<<endl;
}
}
return ;
}

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