POJ 3126 Prim Parh

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13905		Accepted: 7841

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

Sample Output

6
7
0

这个题一看知道初始状态，知道末状态，准备用双向BFS来做，但是老是WA，也一直苦于找不到WA数据。所以暂时先用普通的单向BFS来做，回头再用双向的做一做。

这个题只需要把可以达到的质数入队就可以了。注意判重。

/*************************************************************************
	> File Name: Prim_Path_by_bfs.cpp
	> Author: Zhanghaoran0
	> Mail: chiluamnxi@gmail.com
	> Created Time: 2015年07月29日 星期三 08时37分18秒
 ************************************************************************/

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <math.h>
using namespace std;

struct node{
    int x;
    int step;
}q[10000];
bool flag[10000];
int T;
int head, tail;
int start, end;
bool judge_prim(int x){
    if(x == 1 || x == 0)
        return false;
    if(x == 2)
        return true;
    for(int i = 2; i <= (int)sqrt(x); i ++){
        if(x % i == 0)
            return false;
    }
    return true;
}

int bfs(){
    int temp;
    head = 0;
    tail = 1;
    q[head].step = 0;
    q[head].x = start;
    memset(flag, 0, sizeof(flag));
    while(head < tail){
        for(int i = 1; i <= 9; i += 2){
            temp = q[head].x / 10 * 10 + i;
            if(judge_prim(temp) && !flag[temp]){
                if(temp == end)
                    return q[head].step + 1;
                else{
                    q[tail].x = temp;
                    q[tail ++].step = q[head].step + 1;
                    flag[temp] = true;
                }
            }
            else 
               continue;
        }

        for(int i = 0; i <= 9; i ++){
            temp = q[head].x /100 * 100 + 10 * i + q[head].x % 10;
            if(judge_prim(temp) && !flag[temp]){
                if(temp == end)
                    return q[head].step + 1;
                else{
                    q[tail].x = temp;
                    q[tail ++].step = q[head].step + 1;
                    flag[temp] = true;
                }
            }
            else 
                continue;
        }

        for(int i = 0; i <= 9; i ++){
            temp  = q[head].x / 1000 * 1000 + q[head].x / 10 % 10 * 10 + q[head].x % 10 + i * 100;
            if(judge_prim(temp) && !flag[temp]){
                if(temp == end)
                    return q[head].step + 1;
                else{
                    q[tail].x = temp;
                    q[tail ++].step = q[head].step + 1;
                    flag[temp] = true;
                }
            }
            else 
                continue;
        }

        for(int i = 1; i <= 9; i ++){
            temp = q[head].x % 1000 + i * 1000;
            if(judge_prim(temp) && !flag[temp]){
                if(temp == end)
                    return q[head].step + 1;
                  else{
                  q[tail].x = temp;
                  q[tail ++].step = q[head].step + 1;
                  flag[temp] = true;
                }
            }
        else 
            continue;
        }
        head ++;
    }
    return -1;
}

int main(void){
    cin >> T;
    int steps;
    while(T --){
        cin >> start >> end;
        if(start == end){
            cout << "0" << endl;
            continue;  
         }
        steps = bfs();
        if(steps != -1){
            cout << steps << endl;
            continue;
        }
        else 
            cout << "IMPOSSIBLE" << endl;
    }
    return 0;
}