POJ 3126 Prime Path(BFS)

原题

Prime Path

Time Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

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

题意

从一个素数变化为另一个素数，只能以这样的方式变化：

1.一次之改变某一位上的数字

2.改变之后的数字也必须是素数

求这种改变的次数，若无法达到要求，则输出“Impossible”

涉及知识及算法

素数筛选 BFS深度搜索

代码

#include <iostream>#include <queue>#include <cstring>#include <cstdio>using namespace std;int p[10000];             //素数集合int b[10000];             //标记是否访问过bool pb[10000];           //记录递归深度int n2;queue<int> q;void GetPrime(){    memset(p,0,sizeof(p));    for(int i=2;i<5000;i++)    {        for(int j=2;i*j<10000;j++)        {            //cout<<i<<"*"<<j<<"="<<i*j<<"是合数"<<endl;            p[i*j]=1;                 //标记为合数        }    }}void BFS(int a){    if(a==n2||q.empty()) return;    q.pop();    for(int i=0;i<10;i++)    {        if(p[a-a%10+i]==0&&pb[a-a%10+i]==0)        {            q.push(a-a%10+i);            pb[a-a%10+i]=1;            b[a-a%10+i]=b[a]+1;        }        if(p[a-a/10%10*10+i*10]==0&&pb[a-a/10%10*10+i*10]==0)        {            q.push(a-a/10%10*10+i*10);            pb[a-a/10%10*10+i*10]=1;            b[a-a/10%10*10+i*10]=b[a]+1;        }        if(p[a-a/100%10*100+i*100]==0&&pb[a-a/100%10*100+i*100]==0)        {            q.push(a-a/100%10*100+i*100);            pb[a-a/100%10*100+i*100]=1;            b[a-a/100%10*100+i*100]=b[a]+1;        }        if(i>0&&p[a-a/1000%10*1000+i*1000]==0&&pb[a-a/1000%10*1000+i*1000]==0)        {            q.push(a-a/1000%10*1000+i*1000);            pb[a-a/1000%10*1000+i*1000]=1;            b[a-a/1000%10*1000+i*1000]=b[a]+1;        }    }    BFS(q.front());}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    int kase,n1;    cin >>kase;    GetPrime();    while(kase--)    {        memset(pb,0,sizeof(pb));        memset(b,0,sizeof(b));        cin>>n1>>n2;        q.push(n1);        pb[n1]=1;        BFS(n1);        if(b[n2]||n1==n2)cout<<b[n2]<<endl;        else cout<<"Impossible"<<endl;        while(!q.empty())        {            q.pop();        }    }    return 0;}