Prime Path
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 7 Accepted Submission(s) : 3
Problem 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.
1033The 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.
1733
3733
3739
3779
8779
8179
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
PKU
#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
using namespace std;
int isprime[10005];
struct nnd
{
int first;
int second;
};
int a,e;
int vis[10005];
void dfs(queue<nnd> q)
{
while(!q.empty())
{
nnd t;
nnd r;
t=q.front();
q.pop();
//cout<<e<<endl;
if(t.first==e)
{
//cout<<"hello";
cout<<t.second<<endl;
return ;
}
int k=t.first%1000;
for(int i=1000;i<10000;i=i+1000)
{
if(isprime[k+i]&&!vis[k+i])
{
r.first=k+i; r.second=t.second+1;
q.push(r);
vis[i+k]=1;
}
}
int kk=t.first%100;
int kkk=t.first/1000*1000;
k=kk+kkk;
for(int i=0;i<1000;i=i+100)
{
if(isprime[i+k]&&!vis[i+k])
{
r.first=i+k;
r.second=t.second+1;
q.push(r);
vis[i+k]=1;
}
}
kk=t.first%10;
kkk=t.first/100*100;
k=kk+kkk;
for(int i=0;i<100;i=i+10)
{
if(isprime[i+k]&&!vis[i+k])
{
r.first=i+k;
r.second=t.second+1;
q.push(r);
vis[i+k]=1;
}
}
k=t.first/10*10;
for(int i=1;i<10;i=i+2)
{
if(isprime[i+k]&&!vis[i+k])
{
r.first=i+k;
r.second=t.second+1;
q.push(r);
vis[i+k]=1;
}
}
}
cout<<"Impossible"<<endl;
return;
}
int main()
{
int n;
cin>>n;
for(int i=0;i<10005;i++)
{
isprime=i;
}
for(int i=2;i*i<10010;i++)
{
if(isprime!=0)
for(int j=2;j*i<10010;j++)
{
isprime[i*j]=0;
}
}
isprime[1]=0;
for(int o=0;o<n;o++)
{
queue<nnd> q;
nnd a;
memset(vis,0,sizeof(vis));
cin>>a.first>>e;
a.second=0;
q.push(a);
vis[a.first]=1;
dfs(q);
}
return 0;
}