A - Prime Path（11.1.1）
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Description
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#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
bool num[10001],used[10001];
int s,e;
void prim()//判断素数
{
memset(num,true,sizeof(num));
for(int i=2;i*i<10001;i++)
{
if(num[i])
{
for(int j=i+i;j<10000;j=j+i)
num[j]=false;
}
}
memset(num,false,1000);
}
struct node
{
int x,step;
};
int d[4]={1,10,100,1000};//位数
int bfs(int x)
{
queue<node> que;
node tp,num;
int tx;
num.x=x,num.step=0;
used[x]=false,que.push(num);
while(!que.empty())
{
tp=que.front();
que.pop();
for(int i=0;i<4;i++)
for(int j=0;j<=9;j++)
if(i==3&&j==0)
continue;
else
{
if(j!=(tp.x/d[i]%10))
{
tx=((tp.x/d[i]/10)*10+j)*d[i]+tp.x%d[i];
if(used[tx])
{
used[tx]=false,num.x=tx,num.step=tp.step+1,que.push(num);
if(num.x==e)
return num.step;
}
}
} }
}
int main()
{
int t;
cin>>t;
prim();
while(t--)
{
cin>>s>>e;
memcpy(used,num,10000);
if(s==e)
cout<<"0"<<endl;
else cout<<bfs(s)<<endl;
}
return 0;
}