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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 8179The 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.
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).
One line for each case, either with a number stating the minimal cost or containing the word Impossible.
3
1033 8179 1373 8017 1033 1033
6
7 0
题意：
第一行输入一个值t，有t个测试样例，然后接下来的t行都会有两个数m和n，m和n必须是四位数且都是素数，将m每次只能改变一位数，并且改变一位数之后的数值仍然是素数，要求找出来从m变为n的最少步数。
首先想到的就是广搜，然后每次都分别去改变m的个位，十位，百位，千位；判断改变之后的数要符合是素数并且这个数未曾出现过。直到找到的数等于n为止输出结果，如果无法从m变为n那么输出Impossible。
#include#include #include #include #include using namespace std;int m,n;int vis[100000];int check(int a) //判断此数是否符合为素数并且未出现过{ int flag=0; if(a>=1000&&a<10000&&vis[a]==0) { for(int i=2;i*i<=a;i++) { if(a%i==0) { flag=1; break; } } if(flag==0)return 1; } return 0;}struct node{ int x,step;};void bfs(){ memset(vis,0,sizeof(vis)); queue q; struct node nod={m,0}; q.push(nod); vis[m]=1; while(!q.empty()) { struct node newn=q.front(); q.pop(); if(newn.x==n) { printf("%d\n",newn.step); return ; } //个 for(int i=0;i<=9;i++) { int tx=i+newn.x/10*10; if(check(tx)) { struct node d={tx,newn.step+1}; vis[tx]=1; q.push(d); } } //十 for(int i=0;i<=9;i++) { int tx=newn.x%10+i*10+newn.x/100*100; if(check(tx)) { struct node d={tx,newn.step+1}; vis[tx]=1; q.push(d); } } //百 for(int i=0;i<=9;i++) { int tx=newn.x%100+i*100+newn.x/1000*1000; if(check(tx)) { struct node d={tx,newn.step+1}; vis[tx]=1; q.push(d); } } //千 for(int i=1;i<=9;i++) { int tx=newn.x%1000+i*1000; if(check(tx)) { struct node d={tx,newn.step+1}; vis[tx]=1; q.push(d); } } } cout<<"Impossible"<
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