Prime Path （poj 3126 bfs）

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Prime Path

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11703 Accepted: 6640

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

31033 81791373 80171033 1033

Sample Output

670

Source

Northwestern Europe 2006

题意：  给定两个素数（四位数），求第一个数经过几次转换能够得到第二个素数。转换方式：是变换数中某一位的数字（第一位不能为零，其他的变换数字是0~~9），变换之后的数也为素数。

思路：bfs，搜索求最短路径，很容易就想到广度优先搜索；因为广度优先搜索，第一次搜到得到的步数就是最少的步数。另外打素数表提高判断的时候的效率。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define maxn 10005#define MAXN 2005#define mod 1000000009#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6typedef long long ll;using namespace std;bool ISprime[maxn];bool visit[maxn];int m,n,a,b,c,d;struct Node{    int p[4];//用数组存各位数    int step;};void prime()  //素数筛法{    for (int i=2;i<maxn;i++)    {        if (i%2)            ISprime[i]=true;        else            ISprime[i]=false;    }    int m=sqrt(10010.0);    for (int i=3;i<m;i++)    {        if (ISprime[i])        {            for (int j=i+i;j<maxn;j+=i)                ISprime[j]=false;        }    }}int bfs(){    Node st,now;    memset(visit,false,sizeof(visit));    queue<Node>Q;    while (!Q.empty())        Q.pop();    visit[m]=true;    st.p[0]=m/1000;st.p[1]=(m/100)%10;st.p[2]=(m/10)%10;st.p[3]=m%10;//    printf("%d %d %d %d\n",st.a[0],st.a[1],st.a[2],st.a[3]);    st.step=0;    Q.push(st);    while (!Q.empty())    {        st=Q.front();        Q.pop();        if (st.p[0]==a&& st.p[1]==b&&st.p[2]==c&&st.p[3]==d)        {            return st.step;        }        for (int i=0;i<=3;i++)        {            for (int j=0;j<10;j++)            {                if (st.p[i]==j)                    continue;                if (i==0&&j==0)                    continue;                now.p[0]=st.p[0];                now.p[1]=st.p[1];                now.p[2]=st.p[2];                now.p[3]=st.p[3];                now.p[i]=j;                int x=now.p[0]*1000+now.p[1]*100+now.p[2]*10+now.p[3];                if (ISprime[x]&&!visit[x])                {                    visit[x]=true;                    now.step=st.step+1;                    Q.push(now);                }            }        }    }    return -1;}int main(){    prime();    int cas;    scanf("%d",&cas);    while (cas--)    {        scanf("%d%d",&m,&n);        a=n/1000;b=(n/100)%10;c=(n/10)%10;d=n%10;//        printf("%d %d %d %d\n",a,b,c,d);        int ans=bfs();        if (ans==-1)            printf("Impossible\n");        else            printf("%d\n",ans);    }    return 0;}/*31033 81791373 80171033 1033*/