POJ 3126Prime Path(bfs)

Prime Path

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 22398 Accepted: 12401

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

这道题的题意是：给两个四位数的素数n,m。n每次只可以把自己四位数的某一位改变，改变成的数也必须是素数。不能有前导0，因为必须是四位数。n不是只能变大，也可以变小的。求n最少要经过几次变化能变成m。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<queue>using namespace std;#define MAX_N 10000int prime[MAX_N] = {0};int v[MAX_N];struct path{int x;int t;//最短次数};int main(){for (int i = 2; i*i < MAX_N; i++)//素数打表{if (prime[i] == 0){for (int j = 2; i*j < MAX_N; j++){prime[i*j] = 1;}}}int t;cin >> t;while (t--){int n, m;while (cin >> n >> m){queue<path>q;              //初始化memset(v, 0, sizeof(v));path start;start.x = n;start.t = 0;q.push(start);v[n] = 1;while (!q.empty())     //BFS过程{path temp = q.front();if (temp.x == m) { cout << temp.t << endl; break; }int a[4],b[4];for (int i = 0; i < 4; i++){b[i]=a[i] = q.front().x % 10;q.front().x /= 10;}temp.t++;for (int i = 0; i < 4; i++)//通过数组使它只变一位{for (int j = 0; j < 10; j++){b[i] = j;int c = b[0] + b[1] * 10 + b[2] * 100 + b[3] * 1000;if (b[3] == 0)continue;//去前导0if (prime[c] == 0&&v[c]==0)//是素数而且没选过 进入队列里{v[c] = 1;path t = temp;//必须要用新的变量  因为queue里传递的是引用t.x = c;q.push(t);}}b[i] = a[i];      //改完后要复原，保证只改一次}q.pop();   }}}return 0;}