#include<iostream>
#include<queue>
#include<math.h>
using namespace std;
const int MAX=10000;
int visited[10000];
int result[10000];
bool isprimer(int npp)
{
for (int q = 2; q <= sqrt(double(npp)); ++q)
{
if (npp % q == 0)
{
return 0;
}
}
return 1;
}
int BFS(int start,int end)
{
queue<int> q;
if (start == end)
return 0;
q.push(start);
visited[start]=1;
result[start]=0;
while(!q.empty())
{
int temp=q.front();
q.pop();
int next,j,k;
int num[3];
for (int i =0;i<4;++i)
{
num[0]=temp%10;
num[1]=(temp%100)/10;
num[2]=(temp%1000)/100;
num[3]=temp/1000;
for(j=0;j<=9;j++)
{
num[i]=j;
next=num[0]+num[1]*10+num[2]*100+num[3]*1000;
if((next>1000&&visited[next]!=1)&&isprimer(next))
{
q.push(next);
result[next]=result[temp]+1;
visited[next]=1;
}
if(next==end)
{
return result[next];
}
}
}
}
return MAX;
}
int main() {
//freopen("s.txt","r",stdin);
//freopen("key.txt","w",stdout);
int n,k,j,m;
cin>>j;
while(j--)
{
memset(visited,0,sizeof(visited));
memset(result,0,sizeof(result));
cin>>n>>k;
m=BFS(n,k);
if(m!=MAX)
cout<<m<<endl;
else
cout<<"Impossible"<<endl;
}
return 0;
}
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
3
1033 8179
1373 8017
1033 1033
Sample Output
6
7
0
代碼完全可以優化，比如定義一個結構體可以省掉result數組！
懶得改了
發現poj比joj更驚心動魄?。〗洺ompling,結果半天才出現！
posted on 2009-05-13 23:32
luis 閱讀(707)
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