Prime Path poj bfs 3126

Prime Path

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 18284 Accepted: 10291

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

#include<cstdio>#include<cstring>#include<iostream>#include<queue>using namespace std;const int Max=10100;int prime[Max];int vis[Max];int judge,sum;struct node{int num,step;};queue<node>q;node a,b;void bfs(){   q.push(a);   while(!q.empty())   {     node p=q.front();     q.pop();     if(p.num<=1000||vis[p.num]||prime[p.num]) continue;     vis[p.num]=1;     if(p.num==b.num)     {       judge=1;sum=p.step;return;     }     for(int i=1;i<10;i++)     {       node t;       t.num=(p.num+1000*i)%10000;       t.step=p.step+1;       q.push(t);       t.num=(p.num/1000)*1000+(p.num%1000+100*i)%1000;       q.push(t);       t.num=(p.num/100)*100+(p.num%100+10*i)%100;       q.push(t);       t.num=(p.num/10)*10+(p.num%10+i)%10;       q.push(t);     }   }}int main(){  prime[0]=prime[1]=0;  for(int i=2;i<10001;i++)  {    if(prime[i]==0)    {      for(int j=i;j*i<10001;j++)      prime[i*j]=1;    }  }  int t;  cin>>t;  while(t--)  {     sum=judge=0;      while(!q.empty()) q.pop();     memset(vis,0,sizeof(vis));     cin>>a.num>>b.num;     if(a.num==b.num)     {      printf("0\n");continue;     }     bfs();     if(judge)     {        printf("%d\n",sum);     }     else     {      printf("Impossible\n");     }  }}

    思路：用BFS暴搜，每次改变一位数如果是素数就进入队列，
    但是暴搜的前提要把1w以内的素数筛出来。为了方便对每一位上的数的操作，
    本人用的字符串来储存4位数字。

#include <iostream>#include<queue>#include<cstring>#include<cstdio>using namespace std;int prime[100005];int flag[100004];struct node{char str[6];int step;};void getprime(){  for(int i=2;i<=10000;i++)  {    if(prime[i]==0)    {      for(int j=i;j*i<10001;j++)      prime[i*j]=1;    }  }}int bfs(char *x,char *y){  if(strcmp(x,y)==0) return 0;  queue<node>q;  int i,j,num,step;  char ch;  node t;  strcpy(t.str,x);  t.step=0;  num=(t.str[0]-'0')*1000+(t.str[1]-'0')*100+(t.str[2]-'0')*10+(t.str[3]-'0');  flag[num]=1;  q.push(t);  while(!q.empty())  {   t=q.front();   step=t.step;    q.pop();    for(i=0;i<4;i++)    {      ch=t.str[i];      i==0?j='1':j='0';      for(;j<='9';j++)      {         if(j==ch) continue;         t.str[i]=j;         t.step=step+1;         if(strcmp(t.str,y)==0) return t.step;          num=(t.str[0]-'0')*1000+(t.str[1]-'0')*100+(t.str[2]-'0')*10+(t.str[3]-'0');          if(!prime[num]&&!flag[num])          {           flag[num]=1;           q.push(t);          }      }      t.str[i]=ch;    }  }  return -1;}int main(){    getprime();    int t,step;    char x[6],y[6];    cin>>t;    while(t--)    {       memset(flag,0,sizeof(flag));       cin>>x>>y;       step=bfs(x,y);       if(step!=-1)       printf("%d\n",step);       else        printf("Impossible\n");    }    return 0;}