poj3126 Prime Path 简单BFS

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本文介绍了一个有趣的算法问题：寻找两个四位质数之间的最短转换路径，每次只改变一位数字，并保持结果为质数。通过使用BFS搜索算法，文章提供了一种有效的解决方案。

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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.

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
3
1033 8179
1373 8017
1033 1033

Sample Output
6
7
0

题目大意
给定两个四位质数A,B。求从A置换到B的最小步数。（置换：由一个四位质数改变任意一位上的一个数，使其成为另一个四位质数）

输入
第一行一个整数t，接下来t行，每行两个整数A,B。

输出
共t行，每行一个整数，代表最少步数

这个题用BFS搜索，可以先预处理1-9999的质数
注意多组数据清0

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<string>
#include<vector>
#include<stack>
#include<set>
#include<queue>
using namespace std;
const int maxn=10000;
int n,m;
bool p[maxn],p1[maxn];
void work(){
    int i,j,k;
    for(i=2;i<=9999;i++)
        if(!p[i]){
            k=9999/i;
            for(j=2;j<=k;j++)
                p[j*i]=1;
        }
}
struct node{
    int x,y;
};
node make(int x,int y){
    node z;
    z.x=x;
    z.y=y;
    return z;
}
int bfs(){
    queue <node> q;
    int i;
    q.push(make(n,0));
    while(!q.empty()){
        node to=q.front();
        q.pop();
        if(to.x==m)
            return to.y;
        for(i=1;i<=9;i+=2)
            if(!p[(to.x/10)*10+i] && !p1[(to.x/10)*10+i])
                p1[(to.x/10)*10+i]=1,q.push(make((to.x/10)*10+i,to.y+1));
        for(i=0;i<=9;i++)                      
            if(!p[((to.x/100)*100+to.x%10)+i*10] && !p1[((to.x/100)*100+to.x%10)+i*10])                            
                p1[((to.x/100)*100+to.x%10)+i*10]=1,q.push(make(((to.x/100)*100+to.x%10)+i*10,to.y+1));
        for(i=0;i<=9;i++)                      
            if(!p[((to.x/1000)*1000+to.x%100)+i*100] && !p1[((to.x/1000)*1000+to.x%100)+i*100])                            
                p1[((to.x/1000)*1000+to.x%100)+i*100]=1,q.push(make(((to.x/1000)*1000+to.x%100)+i*100,to.y+1));
        for(i=1;i<=9;i++)                      
            if(!p[(to.x%1000)+i*1000] && !p1[(to.x%1000)+i*1000])                            
                p1[(to.x%1000)+i*1000]=1,q.push(make(to.x%1000+i*1000,to.y+1));
    }
    return -1;
}
int main(){  
    int i,j,k;
    work();
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        int ans=bfs();
        if(ans!=-1)
            printf("%d\n",ans);
        else puts("Impossible");
        memset(p1,0,sizeof(p1));
    }
    return 0;
}