【代码超详解】POJ 3126 Prime Path（素数筛+BFS）

Prime Path

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

寻找从 素数n 到素数 m 每次只能变化一位数字的 最少变化次数，且每次变化得到的数字也为素数

题意：

寻找从素数 n 变换到 素数 m 的最少变换次数，每次只能变换一位数字，且每次变化得到的数字也为素数。

（素数）→（素数）

知识储备：

0x3f3f3f3f的十进制是1061109567，是10⁹ 级别的（和0x7fffffff一个数量级），而一般场合下的数据都是小于10⁹ 的，所以它可以作为无穷大使用而不致出现数据大于无穷大的情形。

素数判断方法

bool isPrime(int n)
{
    if(n==0||n==1)
        return false;
    for(int i=2; i<=sqrt(n); i++)
    {
        if(n%i==0)
            return false;
    }
    return true;
}

           

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<queue>
#include<math.h>
using namespace std;

const int manx=10001;
int n,m;
int vis[manx];
int digit[4]={1000,100,10,1};

bool isPrime(int n)//判断素数
{
    if(n==0||n==1)
        return false;
    for(int i=2; i<=sqrt(n); i++)
    {
        if(n%i==0)
            return false;
    }
    return true;
}

bool can(int x)//判断是否符合条件
{
    if(!isPrime(x)||vis[x]!=-1)return false;//不是质数或者被标记过
    else return true;
}

void BFS(int x)//计算几步变换成m
{
    memset(vis,-1,sizeof(vis));//初始化标记数组vis
    queue<int> q;//定义队列q
    q.push(x);//入队
    vis[x]=0;//标记
    while(!q.empty())//队列不为空，刚开始队列装的是n
    {
        int now=q.front();
        int tx=now;//tx用来保存当前这个数
        q.pop();
        for(int i=3; i>=0; i--)
        {
            int nowdigit=now%10;//分别获取个、十、百、千位数
            now=now/10;
            if(i!=0)//不存在前导零，第一位数不是0，需要特殊判断
            {
                for(int j=0; j<10; j++)
                {
                    if(j==nowdigit)
                        continue;
                    else
                    {
                        int newx=(tx-nowdigit*digit[i]+j*digit[i]);
                                 if(can(newx))
                        {
                            vis[newx]=vis[tx]+1;
                            if(newx==m)return;
                            q.push(newx);
                        }

                    }
                }
            }
            else
            {
                for(int j=1; j<10; j++)
                {
                    if(j==nowdigit)
                        continue;
                    else
                    {
                        int newx=(tx-nowdigit*digit[i]+j*digit[i]);
                                 if(can(newx))
                        {
                            vis[newx]=vis[tx]+1;
                            if(newx==m)return;
                            q.push(newx);
                        }

                    }
                }
            }
        }

    }
}


int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        scanf("%d %d",&n,&m);
        BFS(n);
        if(vis[m]==-1)cout<<"Impossible"<<endl;
        else
            cout<<vis[m]<<endl;
    }
    return 0;
}