POJ 3126 Prime Path（BFS）

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

Output

Sample Input

3
1033 8179
1373 8017
1033 1033

Sample Output

6
7
0

题意：有一个素数主义者要换四位数的门牌，给出旧门牌号和新门牌号，每次变换只能换一位数，而且他不与许看到不是素数的四位数出现，求最少经过多少次变换可以让旧门牌号换成新门牌号。

思路：BFS，每次变换只能换一位数，所以要对每一位数字分别处理，而且要记得标记，入队的书就标记上，不要再次入队。只有素数才入队，所以每次入队前判断是不是素数消耗太大，直接打一个素数表，1~10000即可，筛法保持标记形式的状态，不是素数就标记成0即可。

//612 KB	0 ms	
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int _,flag;
int a,b;
int prime[10010];
bool book[10010];
queue<pair<int,int> >que;
void ini()
{
    flag=0;
    memset(book,0,sizeof(book));
    scanf("%d%d",&a,&b);
    while(!que.empty()) que.pop();
}
void bfs()
{
   que.push(make_pair(a,0) );
   book[a]=1;
   while(!que.empty()){
        pair<int,int> t=que.front();
        que.pop();
        if(t.first==b) {printf("%d\n",t.second);flag=1;return ;}
        int x=t.first;
        for(int i=1;i<=9;i++){
            int nx=x%1000+i*1000;//只变换千位数
            if(prime[nx]&&!book[nx]) {que.push(make_pair(nx,t.second+1)); book[nx]=1;}
        }
        for(int i=0;i<=9;i++){
            int nx=x%100+1000*(x/1000)+i*100; //只变换百位数
            if(prime[nx]&&!book[nx]) {que.push(make_pair(nx,t.second+1)); book[nx]=1;}
        }
        for(int i=0;i<=9;i++){
            int nx=x%10+100*(x/100)+i*10; //只变换十位数
            if(prime[nx]&&!book[nx]) {que.push(make_pair(nx,t.second+1)); book[nx]=1;}
        }
        for(int i=0;i<=9;i++){
            int nx=10*(x/10)+i;//只变换个位数
            if(prime[nx]&&!book[nx]) {que.push(make_pair(nx,t.second+1)); book[nx]=1;}
        }
   }
}

int main()
{
    memset(prime,-1,sizeof(prime));
    /*素数筛*/
    for(int i=2;i<=10;i++)  //剔除10以内的全部非素数
        for(int j=2;j*j<=i;j++)
            if(i%j==0) prime[i]=0;
    prime[1]=0;
    
    for(int i=1;i<=10;i++)   //剔除100以内的全部非素数，顺便剔除掉100~10000之间的一些非素数
        if(prime[i])
        for(int j=2;i*j<=10000;j++)
        prime[i*j]=0;
        
    for(int i=11;i<=100;i++) //剔除100~10000之间的全部非素数
        if(prime[i])
        for(int j=2;i*j<=10000;j++)
        prime[i*j]=0;
        
    scanf("%d",&_);
    while(_--)
    {
        ini();
        bfs();
        if(!flag) printf("Impossible\n");
    }
    return 0;
}

POJ 3126 Prime Path（BFS）

原文：http://blog.csdn.net/kalilili/article/details/42979379