POJ-3126-Prime Path-素数-BFS

Prime Path

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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：将一个素数变成另一个素数，且每次只能变一个数字，而且变成的数字也必须是素数，问最少要多少步
解题思路：首先是判断素数，可以用埃式筛法做一个素数表，方便查询，然后可以用BFS求最小步数

AC代码：
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cctype>
#include<cmath>
#include<cstring>
#include<sstream>
#include<set>
#include<map>
#include<queue>
#define maxn 10005
typedef long long ll;
using namespace std;
typedef pair<int,int> P;
int prime[maxn],ans,n;
bool is_prime[maxn];
int main()
{
    ans=0;
    for(int i=0; i<=maxn; i++) is_prime[i]=true;
    is_prime[0]=is_prime[1]=false;
    for(int i=2; i<=maxn; i++)   //埃式筛法
    {
        if(is_prime[i])
        {
            prime[ans++]=i;    //可以顺便把所有素数保存在数组中，本题其实不需要
            for(int j=i*2; j<=maxn; j+=i) is_prime[j]=false;
        }
    }
    cin>>n;
    while(n--)
    {
        int a,b,sum;
        scanf("%d%d",&a,&b);
        queue<P> que;     //pair函数和队列的连用
        sum=0;
        que.push(P(a,sum));   //保存当前数字和步数
        int vis[10000];
        memset(vis,0,sizeof(vis));
        while(!que.empty())
        {
            P x=que.front();
            if(x.first==b)
                break;
            que.pop();
            vis[x.first]=1;   //标记用过的数字，防止出现死循环
            for(int i=0;i<=9;i++)    //改变个位数字
            {
                int y=x.first/10*10+i;
                if(is_prime[y]&&!vis[y])
                    que.push(P(y,x.second+1));
            }
            for(int i=0;i<=9;i++)  //十位
            {
                int y=x.first%10+x.first/100*100+i*10;
                if(is_prime[y]&&!vis[y])
                    que.push(P(y,x.second+1));
            }
            for(int i=0;i<=9;i++)  //百位
            {
                int y=x.first%10+x.first%100/10*10+x.first/1000*1000+i*100;
                if(is_prime[y]&&!vis[y])
                    que.push(P(y,x.second+1));
            }
            for(int i=1;i<=9;i++)   //千位，注意不能为0
            {
                int y=x.first%1000+i*1000;
                if(is_prime[y]&&!vis[y])
                    que.push(P(y,x.second+1));
            }
        }
        if(que.empty())
            printf("Impossible\n");
        else
            printf("%d\n",que.front().second);
    }
    return 0;
}

**如果有疑问或更好的方法欢迎大家留言（￣︶￣）↗　**