POJ 3126 Prime Path (Guang search)

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12974		Accepted: 7342

Description

The ministers of the Cabinet were quite upset by the message from the chief of Security stating, they would all has T o Change the Four-digit-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 had 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'll just has to paste four new digits over the four old ones on your office door.
-no, it ' s not so simple. Suppose that I change the first digit to an 8 and then the number would read 8033 which is not a prime!
-I see, being the Prime Minister you cannot stand have 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 the prime numbers where only one digit are 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 then the price of a digit is one pound.
-HMM, in this case I need a computer program to minimize the cost. You don ' t know some very cheap software gurus?
-in fact, I do. You see, there are this programming contest going on ... Help the Prime minister to find the cheapest prime path between any and given Four-digit primes! The first digit must be nonzero, of course. Here's 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 step 2 can is reused in the last Step–a new 1 must is purchased.

Input

One line with a positive number:the number of test cases (at most 100). Then for each test case, one line with the numbers separated by a blank. Both numbers is four-digit primes (without leading zeros).

Output

Either with a number stating the minimal cost or containing the word impossible.

Sample Input

31033 81791373 80171033 1033

Sample Output

670

the main topic: to two primes A, B (is 4 digits), asked whether a can be transformed, to change the principle: only one digit of a at a time, and the converted number must be a prime number, such as 1033 can become 1733, but not become 1233 (because 1233 is not a prime), Also can not become 3733 (because from 1033 to 37,331 times the transformation of 2 digits), if possible, the output of the minimum number of transformations, otherwise output impossible.
wide Search, judging whether it is a prime number can be first hit the table.

#include <stdio.h> #include <queue> #include <string.h>using namespace std; #define HUR 100#define THO 1000#define TEN 10const int maxn=10000;bool prime[maxn+5];int vis[maxn],s,e;void prime_table () {int I,j;memset (prime,0, sizeof (prime)); for (i=2;i<maxn;i++) if (!prime[i]) for (j=i*i;j<maxn;j+=i) prime[j]=1;} int BFS () {int I;memset (vis,0,sizeof (VIS));vis[s]=1;queue<int> Que;que.push (s); while (!que.empty ()) {int t= Que.front (); Que.pop (); int d=t;d%=1000;for (i=1;i<10;i++) {int tt=d+i*tho;//transform thousand if (prime[tt]==0 && Vis[tt] ==0) {if (tt==e) return Vis[t];que.push (TT); vis[tt]=vis[t]+1;}} d=t%100+ (t/1000*1000); for (i=0;i<10;i++) {int tt=d+i*hur;//transform hundred if (prime[tt]==0 && vis[tt]==0) {if (tt==e) Return Vis[t];que.push (TT); vis[tt]=vis[t]+1;}} D=t%10+t/100*100;for (i=0;i<10;i++) {int tt=d+i*ten;//transform 10-bit if (prime[tt]==0 && vis[tt]==0) {if (tt==e) return Vis[t];que.push (TT); vis[tt]=vis[t]+1;}} D=t/10*10;for (i=0;i<10;i++) {int tt=d+i;//transform digit if (prime[tt]==0 && vis[tt]==0) {if (tt==e) return Vis[t];que.push (TT); vis[tt]=vis[t]+1;}}} return 0;} int main () {int t,res;prime_table (), scanf ("%d", &t), while (t--) {scanf ("%d%d", &s,&e), if (s==e) {printf ("0\ n "); continue;} Res=bfs (); if (res==0) printf ("impossible\n"); elseprintf ("%d\n", res);} return 0;}

POJ 3126 Prime Path (Guang search)