poj3126 Prime Path 广搜bfs

题目：

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

题意：

      问从a最少需要改变几次才能变成b，每次只能改变一位数字，且改变后的数字必须为素数。

题解：

     用广搜来解决，每次都从千位到各位依次改变一位判断是否满足题目要求，满足即加入队列。

代码：

#include <iostream>
#include <stdio.h>
#include <queue>
#include <math.h>
#include <string.h>
using namespace std;
int a,b;
int turn[4];          //将每次从队列中取出的数转化为数组的形式，便于改变每一位的数字
bool isprime[10003];   //打表，将素数记录下来
int visited[10000];   //判断某一数字是否出现过
int step[10000];      //判断每一位数字从初始状态到此翻转了几次
queue<int>q;

void prime()
{
    int i,j;
    for(i=1000;i<=10000;i++){
        for(j=2;j<i;j++) 
                    if(i%j==0)
                        {isprime[i]=false;break;}
        if(j==i) isprime[i]=true;
    }
}

void turned(int u)
{
    int i;
    for(i=3;i>=0;i--)
    {
        turn[i]=u%10;
        u/=10;
    }
}

int bfs()
{
    int u,i,j;
    q.push(a);
    visited[a]=1;
    while(!q.empty())
    {
        u=q.front();
        q.pop();
        if(u==b) return step[u];
        turned(u);          //将数字转换成数组形式
        for(i=0;i<4;i++)    //依次遍历千百十个位
        {
            int x=turn[i];   //x用来还原数组的数
            for(j=0;j<=9;j++)
            {
                if(i==0&&j==0) continue;
                if(j==x) continue;
                turn[i]=j;
                int v=turn[0]*1000+turn[1]*100+turn[2]*10+turn[3];  //v表示转换后的数
                if(isprime[v]&&!visited[v])
                {
                    step[v]=step[u]+1;
                    visited[v]=1;
                    q.push(v);
                    if(v==b) return step[v];
                }
                turn[i]=x;    //还原
            }
        }
    }
    return -1;
}

int main()
{
    int n;
    prime();
    cin>>n;
    while(n--)
    {
        cin>>a>>b;
        memset(turn,0,sizeof(turn));
        memset(visited,0,sizeof(visited));
        memset(step,0,sizeof(step));
        while(!q.empty())
            q.pop();
        int ans=bfs();
        if(ans==-1) printf("Impossible
");
        else cout<<ans<<endl;
    }
    return 0;
}