POJ 3126 Prime Path（BFS）

Prime Path

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12060 Accepted: 6843

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

31033 81791373 80171033 1033

Sample Output

670

传送门

题意：有一个素数主义者要换四位数的门牌，给出旧门牌号和新门牌号，每次变换只能换一位数，而且他不与许看到不是素数的四位数出现，求最少经过多少次变换可以让旧门牌号换成新门牌号。

思路：BFS，每次变换只能换一位数，所以要对每一位数字分别处理，而且要记得标记，入队的书就标记上，不要再次入队。只有素数才入队，所以每次入队前判断是不是素数消耗太大，直接打一个素数表，1~10000即可，筛法保持标记形式的状态，不是素数就标记成0即可。

//612 KB0 ms#include<cstdio>#include<cstring>#include<queue>using namespace std;int _,flag;int a,b;int prime[10010];bool book[10010];queue<pair<int,int> >que;void ini(){    flag=0;    memset(book,0,sizeof(book));    scanf("%d%d",&a,&b);    while(!que.empty()) que.pop();}void bfs(){   que.push(make_pair(a,0) );   book[a]=1;   while(!que.empty()){        pair<int,int> t=que.front();        que.pop();        if(t.first==b) {printf("%d\n",t.second);flag=1;return ;}        int x=t.first;        for(int i=1;i<=9;i++){            int nx=x%1000+i*1000;//只变换千位数            if(prime[nx]&&!book[nx]) {que.push(make_pair(nx,t.second+1)); book[nx]=1;}        }        for(int i=0;i<=9;i++){            int nx=x%100+1000*(x/1000)+i*100; //只变换百位数            if(prime[nx]&&!book[nx]) {que.push(make_pair(nx,t.second+1)); book[nx]=1;}        }        for(int i=0;i<=9;i++){            int nx=x%10+100*(x/100)+i*10; //只变换十位数            if(prime[nx]&&!book[nx]) {que.push(make_pair(nx,t.second+1)); book[nx]=1;}        }        for(int i=0;i<=9;i++){            int nx=10*(x/10)+i;//只变换个位数            if(prime[nx]&&!book[nx]) {que.push(make_pair(nx,t.second+1)); book[nx]=1;}        }   }}int main(){    memset(prime,-1,sizeof(prime));    /*素数筛*/    for(int i=2;i<=10;i++)  //剔除10以内的全部非素数        for(int j=2;j*j<=i;j++)            if(i%j==0) prime[i]=0;    prime[1]=0;        for(int i=1;i<=10;i++)   //剔除100以内的全部非素数，顺便剔除掉100~10000之间的一些非素数        if(prime[i])        for(int j=2;i*j<=10000;j++)        prime[i*j]=0;            for(int i=11;i<=100;i++) //剔除100~10000之间的全部非素数        if(prime[i])        for(int j=2;i*j<=10000;j++)        prime[i*j]=0;            scanf("%d",&_);    while(_--)    {        ini();        bfs();        if(!flag) printf("Impossible\n");    }    return 0;}