POJ3126 Prime Path bfs+素数筛选

利用素数筛选判断10000以内的所有素数，将它初始化。

然后就是一个BFS。

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8884		Accepted: 5052

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

3
1033 8179
1373 8017
1033 1033

Sample Output

6
7
0

Source

Northwestern Europe 2006

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;

#define MAXN 10010

struct node
{
    int x,s;
};

int p[MAXN];
bool vis[MAXN];
//素数筛选
void initPrime(int n)
{
    int m=sqrt(n+0.5);
    p[0]=p[1]=0;
    memset(p,0,sizeof(p));
    for(int i=2;i<=m;i++) if(!p[i])
        for(int j=i*i;j<=n;j+=i) p[j]=1;
}

int change(int x,int i,int j)
{
    if(i==1)
        return (x/10)*10+j;
    if(i==2)
        return (x/100)*100+x%10+j*10;
    if(i==3)
        return (x/1000)*1000+x%100+j*100;
    return x%1000+j*1000;
}

bool jud(int x)
{
    if(p[x]) return false;
    if(vis[x]) return false;
    return true;
}

int bfs(int st,int ed)
{
    queue<node> q;
    node a,b;
    a.x=st;
    a.s=0;
    q.push(a);
    vis[st]=1;
    while(!q.empty())
    {
        a=q.front();
        q.pop();
        if(a.x==ed)
            return a.s;
        for(int i=1;i<=4;i++)
        {
            for(int j=0;j<=9;j++)
            {
                if(i==4 && j==0) continue;
                b.x=change(a.x,i,j);
                b.s=a.s+1;
                if(!jud(b.x)) continue;
                vis[b.x]=1;
                q.push(b);
            }
        }
    }
    return -1;
}


int main()
{
    initPrime(10000);
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int st,ed;
        memset(vis,0,sizeof(vis));
        scanf("%d%d",&st,&ed);
        int ans=bfs(st,ed);
        if(ans==-1)
            printf("Impossible\n");
        else
            printf("%d\n",ans);
    }
    return 0;
}