Prime Path
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9982 | Accepted: 5724 |
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
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 100001
const int inf=0x7fffffff; //无限大
const int MAXN = 10000;
bool flag[MAXN];
int primes[MAXN], pi;
struct point
{
int x;
int y;
};
void GetPrime_1()
{
int i, j;
pi = 0;
memset(flag, false, sizeof(flag));
for (i = 2; i < MAXN; i++)
if (!flag[i])
{
primes[i] = 1;//素数标识为1
for (j = i; j < MAXN; j += i)
flag[j] = true;
}
}
int vis[maxn];
int main()
{
GetPrime_1();
int t;
cin>>t;
while(t--)
{
memset(vis,0,sizeof(vis));
int n,m;
cin>>n>>m;
vis[n]=1;
queue<point> q;
q.push((point){n,0});
int flag1=0;
while(!q.empty())
{
point now=q.front();
if(now.x==m)
{
flag1=now.y;
break;
}
point next;
for(int i=0;i<=9;i++)
{
next.x=now.x/10;
next.x*=10;
next.x+=i;
next.y=now.y+1;
if(next.x<1000||next.x>=10000)
continue;
if(vis[next.x]==1)
continue;
if(next.x==m)
{
flag1=next.y;
break;
}
if(primes[next.x]==1)
{
//cout<<next.x<<endl;
vis[next.x]=1;
q.push(next);
}
}
for(int i=0;i<=9;i++)
{
int temp=now.x%10;
next.x=now.x/100;
next.x*=100;
next.x+=i*10;
next.x+=temp;
if(next.x<1000||next.x>=10000)
continue;
if(vis[next.x]==1)
continue;
if(next.x==m)
{
flag1=next.y;
break;
}
if(primes[next.x]==1)
{
//cout<<next.x<<endl;
vis[next.x]=1;
q.push((point){next.x,now.y+1});
}
}
for(int i=0;i<=9;i++)
{
int temp=now.x%100;
next.x=now.x/1000;
next.x*=1000;
next.x+=i*100;
next.x+=temp;
if(next.x<1000||next.x>=10000)
continue;
if(vis[next.x]==1)
continue;
if(next.x==m)
{
flag1=next.y;
break;
}
if(primes[next.x]==1)
{
//cout<<next.x<<endl;
vis[next.x]=1;
q.push((point){next.x,now.y+1});
}
}
for(int i=0;i<=9;i++)
{
int temp=now.x%1000;
next.x=now.x/10000;
next.x*=10000;
next.x+=i*1000;
next.x+=temp;
if(next.x<1000||next.x>=10000)
continue;
if(vis[next.x]==1)
continue;
if(next.x==m)
{
flag1=next.y;
break;
}
if(primes[next.x]==1)
{
// cout<<next.x<<endl;
vis[next.x]=1;
q.push((point){next.x,now.y+1});
}
}
if(flag1>0)
break;
q.pop();
}
printf("%d\n",flag1);
}
return 0;
}