2013
12-26

# Prime Path

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.

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).

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).

3
1033 8179
1373 8017
1033 1033

6
7
0

/*
* File:   hdu1973.cpp
* Author: ssslpk
* Created on August 28, 2012, 4:34 PM
* 题意：给出两个四位数，现要改变第一个数中的个，十，百，千位当中的一个数
* 使它最终变成第二个数，要求这过程中形成的数是素数，问最少的步骤
*
* 题解：素数筛选＋bfs
*/
#include <cstdlib>
#include<iostream>
#include<string.h>
#include<cstdio>
#include<math.h>
#include<queue>
#define maxn 10002

using namespace std;
int ispri[maxn];
int pri[maxn];
int plen;

void prime()
{
memset(ispri,1,sizeof(ispri));
plen=0;
ispri[1]=ispri[0]=0;
for(int i=2;i<maxn;i++)
{
if(ispri[i])
{
pri[plen++]=i;
for(int j=2*i;j<maxn;j+=i)
ispri[j]=0;
}
}
}
struct node
{
int num;
int id;
int cnt;
};
int serch(int num,int id)
{
while(id){num/=10;id--;}
return num%10;
}
queue<node> Q;
int bfs(int s,int e)
{
if(s==e)return 0;
node first;
first.num=s;
first.id=-1;
first.cnt=0;
int vis[maxn];
memset(vis,0,sizeof(vis));
Q.push(first);
vis[s]=1;
while(!Q.empty())
{
node temp=Q.front();
Q.pop();
if(temp.num==e)
{
while(!Q.empty())Q.pop();
return temp.cnt;
}
for(int j=0;j<4;j++)
if(temp.id!=j)
{
int dig=serch(temp.num,j);
for(int i=0;i<10;i++)
if((j != 3 || i != 0)&& dig !=i)
{
node p;
p.num=temp.num -(dig - i)*pow(10,j);
p.id=j;
p.cnt=temp.cnt+1;
if(ispri[p.num]&& !vis[p.num])
{
vis[p.num]=1;
Q.push(p);

}
}
}
}

}

int main(int argc, char** argv) {
prime();
int cas;
scanf("%d",&cas);
while(cas--)
{
int s,e;
scanf("%d%d",&s,&e);
printf("%d\n",bfs(s,e));
}
return 0;
}

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