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
一个四位素数变化成另一个四位素数，每次只能改变一位且改变后的数字也是素数，求最少需要改变多少次。
放在搜索分类下我才知道要用BFS。。。
#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <algorithm>
#define INF 0x3f3f3f3f
#define MAXN 1010
using namespace std;
struct Node
{
int a[4],step;
};
int primer[3000],cnt,vis[10000];
bool flag;
bool judge(int n)
{
for(int i=2; i*i<=n; ++i)
if(n%i==0)
return false;
return true;
}
void bfs(Node start,int m)
{
queue<Node> q;
q.push(start);
while(!q.empty())
{
Node tmp=q.front(),tmp1;
q.pop();
int mm=tmp.a[0]*1000+tmp.a[1]*100+tmp.a[2]*10+tmp.a[3];
//cout<<"mm="<<mm<<endl;
if(mm==m)
{
cout<<tmp.step<<endl;
flag=true;
break;
}
tmp1=tmp;
for(int i=1; i<=9; ++i)
{
tmp1.a[0]=i;
int t=tmp1.a[0]*1000+tmp1.a[1]*100+tmp1.a[2]*10+tmp1.a[3];
if(judge(t)&&!vis[t])
{
//cout<<"t0="<<t<<endl;
vis[t]=1;
tmp1.step=tmp.step+1;
q.push(tmp1);
}
tmp1=tmp;
}
for(int i=0; i<=9; ++i)
{
for(int j=1; j<4; ++j)
{
tmp1.a[j]=i;
int t=tmp1.a[0]*1000+tmp1.a[1]*100+tmp1.a[2]*10+tmp1.a[3];
if(judge(t)&&!vis[t])
{
//cout<<"t="<<t<<endl;
vis[t]=1;
tmp1.step=tmp.step+1;
q.push(tmp1);
}
tmp1=tmp;
}
}
}
}
int main()
{
ios::sync_with_stdio(false);
int t,n,m;
cin>>t;
while(t--)
{
flag=false;
memset(vis,0,sizeof(vis));
cin>>n>>m;
Node start;
start.a[0]=n/1000;
start.a[1]=(n/100)%10;
start.a[2]=(n/10)%10;
start.a[3]=n%10;
start.step=0;
bfs(start,m);
if(!flag)
cout<<"Impossible"<<endl;
}
return 0;
}