D-POJ-3126 Prime Path

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

BFS，将素数打表可以缩短运行时间。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<queue>
#include<cstring>
using namespace std;

#define maxn 10005
bool vis[maxn];          //标记是否访问
bool IsPrime[maxn];      //标记素数

void Prime(){   //素数筛法
IsPrime[0]=IsPrime[1]=false;IsPrime[2]=true;
for(int i=3;i<maxn;i++){
if(i%2)
IsPrime[i]=true;
else
IsPrime[i]=false;
}
int m=sqrt(maxn*1.0);
for(int i=3;i<m;i++){
if(IsPrime[i]){
for(int j=i;j<maxn;j+=i)
IsPrime[j]=false;
}
}
}

struct Node{
int p[4];
int step;
};

void BFS(int a,int b){
memset(vis, false, sizeof(vis));
Node now,next;
int x;
queue<Node>Q;
now.p[0]=a/1000;now.p[1]=a%1000/100;now.p[2]=a%100/10;now.p[3]=a%10;
now.step=0;
Q.push(now);
while(!Q.empty()){
now=Q.front();
Q.pop();
if(now.p[0]*1000+now.p[1]*100+now.p[2]*10+now.p[3]==b){
printf("%d\n",now.step);
return;
}
for(int i=0;i<=3;i++){
for(int j=0;j<=9;j++){
if(i==0&&j==0)
continue;
if(now.p[i]==j)
continue;
next.p[0]=now.p[0];
next.p[1]=now.p[1];
next.p[2]=now.p[2];
next.p[3]=now.p[3];
next.p[i]=j;
x=next.p[0]*1000+next.p[1]*100+next.p[2]*10+next.p[3];
if(!vis[x]&&IsPrime[x]){
next.step=now.step+1;
vis[x]=true;
Q.push(next);
}
}
}
}
printf("Impossible\n");
}

int main(){
Prime();
int n,a,b;
scanf("%d",&n);
while(n--){
scanf("%d%d",&a,&b);
BFS(a,b);
}
return 0;
}

开始不管怎么运行都是输出Impossible，后来发现Prim函数没运行，蠢哭~