poj3126

题目名称：Prime Path

题目链接：http://poj.org/problem?id=3126

 

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

 

 

题意：从一个素数变成另一个素数，每次只能改变其中的一位，而且改变后的值得是素数，求到达的最短次数

代码如下：

 

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
struct Node
{
    string c;        //表示数字
    int sum;         //表示步数
    Node(string c,int sum)
    {
        this->c=c;
        this->sum=sum;
    }
};
string a,b;
int vis[10000];
bool primer[10000];
queue<Node> q;
void bfs()
{
    while(!q.empty())
        q.pop();
    q.push(Node(a,0));
    if(a==b)
    {
        cout<<'0'<<endl;
        return ;
    }
    while(!q.empty())
    {
        Node now=q.front();
        q.pop();
        for(int i=0;i<10;i++)
        {
            for(int j=0;j<4;j++)
            {
                string tmp=now.c;
                tmp[j]=i+'0';
                int num=(tmp[0]-'0')*1000+(tmp[1]-'0')*100+(tmp[2]-'0')*10+tmp[3]-'0';
                if(!vis[num]&&primer[num])
                {
                    vis[num]=true;
                    q.push(Node(tmp,now.sum+1));
                }
                if(tmp==b)
                {
                    cout<<now.sum+1<<endl;
                    return ;
                }
            }
        }

    }
}
int main()
{
    int n;
    memset(primer,true,sizeof(primer));
    for(int i=2;i<=sqrt(10000+0.5);i++)  //素数筛法
    {
        if(primer[i])
            for(int j=i*i;j<=10000;j+=i)
                primer[j]=false;
    }
    for(int i=0;i<1000;i++)
        primer[i]=false;
    while(cin>>n)
    {
        for(int i=0;i<n;i++)
        {
            cin>>a>>b;
            memset(vis,0,sizeof(vis));
            bfs();
        }
    }
    return 0;
}