POJ-3126-宽搜+素数筛选

Prime Path

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11379 Accepted: 6448

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

该题目先打个素数表，因为求最小步数，很自然就想到了宽搜了。该题对该四位数的每一位进行操作。具体看代码：

#include<queue>#include<math.h>#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;const int maxn=10000;void divide(int nn,int mm[]){    int chu=1000;    for(int i=0;i<4;i++)    {        mm[i]=nn/chu;        nn=nn%chu;        chu/=10;    }}int main(){    bool vis[maxn];    memset(vis,true,sizeof(vis));    vis[0]=vis[1]=false;    for(int i=2;i*i<=10000;i++)       //素数筛选    {        if(vis[i])        {            for(int j=i*i;j<=10000;j+=i)                vis[j]=false;        }    }    int t;    scanf("%d\n",&t);    while(t--)    {        int m,n;        scanf("%d%d",&m,&n);        int num[4];        int nnn[4];        int d[maxn];        int dd[maxn];        queue<int>q;        q.push(m);        memset(d,0,sizeof(d));        memset(dd,0,sizeof(dd));        dd[m]=1;        while(q.size())        {            int p=q.front(); q.pop();            if(p==n) break;            divide(p,num);            for(int i=0;i<4;i++)            {                for(int k=0;k<4;k++)                    nnn[k]=num[k];                                    for(int j=(i==0)?1:0;j<10;j++)                {                    if(j==num[i]) continue;                    nnn[i]=j;                    int ss=nnn[0]*1000+nnn[1]*100+nnn[2]*10+nnn[3];                    if(vis[ss]&&dd[ss]==0)                    {                        dd[ss]=1;                        d[ss]=d[p]+1;                        q.push(ss);                    }                }            }        }        int sd=0;        if(d[n]>0 || m==n) sd=1;          if(sd)   printf("%d\n",d[n]);        else     printf("Impossible\n");    }return 0;}