acm训练2019，1，19 POJ - 3126首相的质数

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

#include <iostream>
#include <queue>
using namespace std;
#define MAXV 10000
 
bool prime[MAXV];


 
void init(){    //素数打表
    int i,j;
    for(i=1000;i<=MAXV;i++){
        for(j=2;j<i;j++)
            if(i%j==0){
                prime[i]=false;
                break;
            }
        if(j==i) prime[i]=true;
    }
}
 
int bfs(int first,int last)//
{
    bool dis[MAXV];
    queue <int>q;
    int v,i,j,temp,vtemp,count[MAXV],t[4];
    memset(dis,false,sizeof(dis));
    memset(count,0,sizeof(count));
 
    q.push(first);
    dis[first]=true;
 
    while(!q.empty()){
        v=q.front();
        q.pop();
 
        t[0]=v/1000;
        t[1]=v%1000/100;
        t[2]=v%100/10;
        t[3]=v%10;

 
        for(j=0;j<4;j++){
            temp=t[j];
            for(i=0;i<10;i++)
                if(i!=temp){
                    t[j]=i;
                    vtemp=t[0]*1000+t[1]*100+t[2]*10+t[3];
                    if(!dis[vtemp] && prime[vtemp]){
                        count[vtemp]=count[v]+1;
                        dis[vtemp]=true;
                        q.push(vtemp);
                    }
                    if(vtemp==last) return count[vtemp];
                }
            t[j]=temp;
        }
        if(v==last) return count[v];
    }
    return -1;
}
 
int main(){
    int n,a,b,key;
    init();
    scanf("%d",&n);
    while(n--){
        scanf("%d%d",&a,&b);
        key=bfs(a,b);//入口
        if(key!=-1) printf("%d\n",key);
        else printf("Impossible\n");
    }
    return 0;
}