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技术 2022年11月23日
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Prime Path

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 14091 Accepted: 7959

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

Source

Northwestern Europe 2006

没什么好说的,BFS搜索,就是在搜索的时候,千位不能为零QAQ

#include <map>
#include <list>
#include <climits>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-9
#define LL unsigned long long
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CRR fclose(stdin)
#define CWW fclose(stdout)
#define RR freopen("input.txt","r",stdin)
#define WW freopen("output.txt","w",stdout)const int Max = 10010;
struct node
{
int x;
int num;
};int n,m;
bool prime[Max];
bool vis[Max];
int bfs()
{
memset(vis,false,sizeof(vis));
node a,b;
a.num=0;
a.x=n;
queue<node>Q;
vis[n]=true;
Q.push(a);
while(!Q.empty())
{
a=Q.front();
Q.pop();
if(a.x==m)
{
return a.num;
}
for(int i=0;i<10;i++)
{
b.x=a.x/10*10+i;
b.num=a.num+1;
if(!vis[b.x]&&!prime[b.x])
{
vis[b.x]=true;
Q.push(b);
}
}
for(int i=0;i<10;i++)
{
int s=a.x%10;
b.x=a.x/100*100+i*10+s;
b.num=a.num+1;
if(!vis[b.x]&&!prime[b.x])
{
vis[b.x]=true;
Q.push(b);
}
}
for(int i=0;i<10;i++)
{
int s=a.x%100;
b.x=a.x/1000*1000+i*100+s;
b.num=a.num+1;
if(!vis[b.x]&&!prime[b.x])
{
vis[b.x]=true;
Q.push(b);
}
}
for(int i=1;i<10;i++)
{
b.x=a.x%1000+i*1000;
b.num=a.num+1;
if(!vis[b.x]&&!prime[b.x])
{
vis[b.x]=true;
Q.push(b);
}
}
}
return 0;
}
int main()
{
memset(prime,false,sizeof(prime));
prime[1]=true;
prime[0]=true;
m=sqrt(Max)+1;
for(int i=2;i<m;i++)
{
if(!prime[i])
{
for(int j=i*i;j<=Max;j+=i)
{
prime[j]=true;
}
}
}
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d %d",&n,&m);
printf("%d\n",bfs());
}
return 0;
}

版权声明:本文为博主原创文章,未经博主允许不得转载。

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