Prime Path
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 7550 | Accepted: 4281 |
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.
1033The 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.
1733
3733
3739
3779
8779
8179
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
简单的BFS题，分别改变一个四位数的个、十、百、千位即可：
代码：
View Code
1 #include <stdio.h> 2 #include <iostream> 3 #include <string.h> 4 using namespace std; 5 6 7 struct M{ 8 int num; 9 int step; 10 }; 11 12 bool prime(int x) 13 { 14 int i,j; 15 for(i=2;i*i<=x;i++) 16 { 17 if(x%i==0) 18 return false; 19 } 20 return true; 21 } 22 23 int BFS(int a, int b) 24 { 25 M queue[100000]; 26 bool vis[100000]={false}; 27 int i,j; 28 int head,tail; 29 head=tail=0; 30 queue[0].num=a; 31 queue[tail++].step=0; 32 while(head<tail) 33 { 34 M x=queue[head++]; 35 if(x.num==b) 36 { 37 return x.step; 38 } 39 int aft; 40 for(i=1;i<=9;i++) //改变个位 41 { 42 if(!vis[x.num/10*10+i] && prime(x.num/10*10+i)==true) 43 { 44 vis[x.num/10*10+i]=true; 45 queue[tail].num=x.num/10*10+i; 46 queue[tail++].step=x.step+1; 47 //cout<<x.num/10*10+i<<"....."<<endl; 48 } 49 } 50 for(i=0;i<=9;i++) //改变十位 51 { 52 aft=x.num%10+x.num/100*100+i*10; 53 if(!vis[aft] && prime(aft)) 54 { 55 vis[aft]=true; 56 queue[tail].num=aft; 57 queue[tail++].step=x.step+1; 58 //cout<<aft<<"......"<<endl; 59 } 60 } 61 for(i=0;i<=9;i++) //改变百位 62 { 63 aft=x.num%100+x.num/1000*1000+i*100; 64 if(!vis[aft] && prime(aft)) 65 { 66 vis[aft]=true; 67 queue[tail].num=aft; 68 queue[tail++].step=x.step+1; 69 //cout<<aft<<"....."<<endl; 70 } 71 } 72 for(i=1;i<=9;i++) //改变千位 73 { 74 aft=x.num%1000+i*1000; 75 if(!vis[aft] && prime(aft)) 76 { 77 vis[aft]=true; 78 queue[tail].num=aft; 79 queue[tail++].step=x.step+1; 80 //cout<<aft<<"......"<<endl; 81 } 82 } 83 } 84 return 0; 85 } 86 87 int main() 88 { 89 int t; 90 int i,j,k; 91 scanf("%d",&t); 92 while(t--) 93 { 94 int n; 95 int m; 96 scanf("%d%d",&n,&m); 97 printf("%d\n",BFS(n,m)); 98 } 99 return 0; 100 }