poj 3126 Prime Path

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

Source

Northwestern Europe 2006

题目大意：

就是给你两个数 a 和 b，让你把 a 变成 b最少需要几步（每次变化 a 与 b 都是素数） 

解题思路：

首先判断一下是不是素数，然后枚举个位十位百位和千位，用bfs搜索：

上代码

/*
2015 - 09 - 09


Author: ITAK


Motto:
今日的我要超越昨日的我，明日的我要胜过今日的我，
以创作出更好的代码为目标，不断地超越自己。
*/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;

const int maxn = 1e4;
const int Max = 15000;

struct node
{
    int prime;
    int step;
}q[Max];

bool vis[Max];
int a, b;

bool isprime(int data)
{
    if(data==2 || data==3)
        return 1;
    if(data<=1 || data%2==0)
        return 0;
    else if(data > 3)
    for(int i=3; i*i<=data; i+=2)
        if(!(data%i))
        return 0;
    return 1;
}

void bfs()
{
    int head, tail;
    head = tail = 0;
    q[head].prime = a;
    q[tail++].step = 0;
    vis[a] = 1;
    while(head < tail)
    {
        node x = q[head++];
        if(x.prime == b)
        {
            cout<<x.step<<endl;
            return;
        }
        int numg = x.prime%10;///个位
        int nums = (x.prime/10)%10;///十位
        ///分别枚举个位 十位 百位 千位
        for(int i=1; i<=9; i+=2)///个位数是偶数肯定不是素数
        {
            int dy = (x.prime/10)*10+i;
            if(dy!=x.prime && !vis[dy] && isprime(dy))
            {
                vis[dy] = 1;
                q[tail].prime = dy;
                q[tail++].step = x.step+1;
            }
        }

        for(int i=0; i<=9; i++)
        {
            int dy = (x.prime/100)*100+i*10+numg;
            if(dy!=x.prime && !vis[dy] && isprime(dy))
            {
                vis[dy] = 1;
                q[tail].prime = dy;
                q[tail++].step = x.step+1;
            }
        }

        for(int i=0; i<=9; i++)
        {
            int dy = (x.prime/1000)*1000+i*100+nums*10+numg;
            if(dy!=x.prime && !vis[dy] && isprime(dy))
            {
                vis[dy] = 1;
                q[tail].prime = dy;
                q[tail++].step = x.step+1;
            }
        }

        for(int i=1; i<=9; i++)///千位必须大于0
        {
            int dy = x.prime%1000+1000*i;
            if(dy!=x.prime && !vis[dy] && isprime(dy))
            {
                vis[dy] = 1;
                q[tail].prime = dy;
                q[tail++].step = x.step+1;
            }
        }
    }

    puts("Impossible");
    return;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        cin>>a>>b;
        memset(vis, 0, sizeof(vis));
        bfs();
    }
    return 0;
}