poj3126Prime Path(bfs水题～～～我喜欢哈哈哈哈哈哈)-优快云博客

本文链接：https://blog.youkuaiyun.com/SDUTyangkun/article/details/52185757

本文介绍了一种算法，用于解决从一个四位质数到另一个四位质数的问题，通过一系列质数转换，每次只改变一位数字，寻找成本最低的转换路径。

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Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16991		Accepted: 9561

Description

The ministers of the cabinet(内阁) 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 m(主要的)inister, you know!
— I know, so therefore your new number 8179 is also a prime. You will just have to paste f(张贴)our new digits o(数字)ver the four old ones on your office door.
— No, it’s not that simple. Suppose that I change the first digit to (数字)an 8, then the number will read 8033 which is not a prime!(初期)
— 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 f(计划)or 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(偷听), intervened(干涉).
— No unnecessary expenditure, (支出)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 t(使减到最少)he cost. You don't know some very cheap software gurus, (古鲁)do you?
— In fact, I do. You see, there is this programming contest g(竞赛)oing 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 i(解决方案)n 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(膏的) over in step 2 can not be reused in the last step – a new 1 must be purchased.(购买)

Input

One line with a positive(积极的) 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(最低的) cost or containing the word Impossible.

Sample Input

Sample Output

6
7
0

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

struct node
{
    int x;
    int step;
}tmp,tmp1;

int vis[10000];
int judge(int x)
{
    int i;
    if(x==1||x==0||x==2)
        return 0;
    else
    {
            for(i=2;i<=sqrt(x);i++)
       {
              if(x%i==0)
              return 0;
       }
       return 1;

    }

}
void work()
{
    int n,m,i,x;
    scanf("%d%d",&n,&m);
    int flag=0;
        vis[n]=1;
        tmp.x=n;
        tmp.step=0;

        queue<node>q;
        q.push(tmp);

        while(!q.empty())
        {
            tmp=q.front();
            q.pop();

            if(tmp.x==m)
            {
                printf("%d\n",tmp.step);
                flag=1;
                break;
            }

            tmp.step++;
            for(i=0;i<=9;i++)
            {
                x=tmp.x/10*10+i;
                if(judge(x)&&!vis[x])
                {
                    tmp1=tmp;
                    tmp1.x=x;
                    vis[x]=1;
                    q.push(tmp1);
                }
            }

            for(i=0;i<=9;i++)
            {
                x=tmp.x/100*100+i*10+tmp.x%10;
                if(judge(x)&&!vis[x])
                {
                    tmp1=tmp;
                    tmp1.x=x;
                    vis[x]=1;
                    q.push(tmp1);
                }
            }

            for(i=0;i<=9;i++)
            {
                x=tmp.x/1000*1000+i*100+tmp.x%100;
                if(judge(x)&&!vis[x])
                {
                    tmp1=tmp;
                    tmp1.x=x;
                    vis[x]=1;
                    q.push(tmp1);
                }
            }

            for(i=1;i<=9;i++)
            {
                x=i*1000+tmp.x%1000;
                if(judge(x)&&!vis[x])
                {
                    tmp1=tmp;
                    tmp1.x=x;
                    vis[x]=1;
                    q.push(tmp1);
                }
            }
        }
        if(!flag) printf("Impossible\n");
}
int main()
{
    int t;
    //int m,n;
    int i,j;

    scanf("%d",&t);
    while(t--)
    {
         memset(vis,0,sizeof(vis));
        work();
    }

    return 0;

}