POJ 3126 Prime Path （BFS）

Prime Path

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9982 Accepted: 5724

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

题意：给定两个素数n和m，要求把n变成m，每次变换时只能变一个数字，即变换后的数与变换前的数只有一个数字不同，并且要保证变换后的四位数也是素数。求最小的变换次数；如果不能完成变换，输出Impossible。

无论怎么变换，个位数字一定是奇数（个位数字为偶数肯定不是素数），这样枚举个位数字时只需枚举奇数就行；而且千位数字不能是0。所以可以用广搜，枚举各个数位上的数字，满足要求的数就加入队列，直到变换成功。因为是广搜，所以一定能保证次数最少。

AC代码：

#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<algorithm>using namespace std;int n, m;const int N = 1e4 + 100;int vis[N];struct node{    int x, step;};queue<node> Q;bool judge_prime(int x) //判断素数{    if(x == 0 || x == 1)        return false;    else if(x == 2 || x == 3)        return true;    else    {        for(int i = 2; i <= (int)sqrt(x); i++)            if(x % i == 0)                return false;        return true;    }}void BFS(){    int X, STEP, i;    while(!Q.empty())    {        node tmp;        tmp = Q.front();        Q.pop();        X = tmp.x;        STEP = tmp.step;        if(X == m)        {            printf("%d\n",STEP);            return ;        }        for(i = 1; i <= 9; i += 2) //个位        {            int s = X / 10 * 10 + i;            if(s != X && !vis[s] && judge_prime(s))            {                vis[s] = 1;                node temp;                temp.x = s;                temp.step = STEP + 1;                Q.push(temp);            }        }        for(i = 0; i <= 9; i++) //十位        {            int s = X / 100 * 100 + i * 10 + X % 10;            if(s != X && !vis[s] && judge_prime(s))            {                vis[s] = 1;                node temp;                temp.x = s;                temp.step = STEP + 1;                Q.push(temp);            }        }        for(i = 0; i <= 9; i++) //百位        {            int s = X / 1000 * 1000 + i * 100 + X % 100;            if(s != X && !vis[s] && judge_prime(s))            {                vis[s] = 1;                node temp;                temp.x = s;                temp.step = STEP + 1;                Q.push(temp);            }        }        for(i = 1; i <= 9; i++) //千位        {            int s = i * 1000 + X % 1000;            if(s != X && !vis[s] && judge_prime(s))            {                vis[s] = 1;                node temp;                temp.x = s;                temp.step = STEP + 1;                Q.push(temp);            }        }    }    printf("Impossible\n");    return ;}int main(){    int t, i;    scanf("%d",&t);    while(t--)    {        while(!Q.empty()) Q.pop();        scanf("%d%d",&n,&m);        memset(vis,0,sizeof(vis));        vis[n] = 1;        node tmp;        tmp.x = n;        tmp.step = 0;        Q.push(tmp);        BFS();    }    return 0;}

有一点我不明白：上面的代码选G++可以AC，但是用C++就Compile Error。