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<stdio.h>
#include<math.h>
#include<string.h>
int a,b,s;
int book[100010];
struct Nodes
{
int x;
int s;
}q[10100];
int p1(int x)
{
int i,k;
k=sqrt(x);
for(i=2;i<=k;i++)
if(x%i==0)
break;
if(i==k+1)
return 1;
else
return 0;
}
void dfs(int a)
{
memset(book,0,sizeof(book));
int head,tail,ge,shi,bai,qian;
head=1;tail=1;
q[tail].x=a;
q[tail].s=0;
tail++;
book[a]=1;
int sum,i,flag=0;
while(head<tail)
{
ge=q[head].x%10;
shi=q[head].x/10%10;
bai=q[head].x/100%10;
qian=q[head].x/1000;
for(i=1;i<=9;i+=2)
{
sum=qian*1000+bai*100+shi*10+i;
if(p1(sum)&&book[sum]==0&&sum<10000)
{
book[sum]=1;
q[tail].x=sum;
q[tail].s=q[head].s+1;
tail++;
if(sum==b)
{
flag=1;
break;
}
}
}
if(flag)
break;
for(i=0;i<=9;i++)
{
sum=qian*1000+bai*100+i*10+ge;
if(p1(sum)&&book[sum]==0&&sum<10000)
{
book[sum]=1;
q[tail].x=sum;
q[tail].s=q[head].s+1;
tail++;
if(sum==b)
{
flag=1;
break;
}
}
}
if(flag)
break;
for(i=0;i<=9;i++)
{
sum=qian*1000+i*100+shi*10+ge;
if(p1(sum)&&book[sum]==0&&sum<10000)
{
book[sum]=1;
q[tail].x=sum;
q[tail].s=q[head].s+1;
tail++;
if(sum==b)
{
flag=1;
break;
}
}
}
if(flag)
break;
for(i=1;i<=9;i++)
{
sum=i*1000+bai*100+shi*10+ge;
if(p1(sum)&&book[sum]==0&&sum<10000)
{
book[sum]=1;
q[tail].x=sum;
q[tail].s=q[head].s+1;
tail++;
if(sum==b)
{
flag=1;
break;
}
}
}
if(flag)
break;
head++;
}
if(!flag)
printf("Impossible\n");
else
printf("%d\n",q[tail-1].s);
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d %d",&a,&b);
if(a==b)
printf("0\n");
else
dfs(a);
}
return 0;
}题意 给你两个4位数，要求你每次只能变换一位数字，并且变换前后的数字都要满足是素数；求最少变换次数
思路 bfs和打表