Prime Path
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 21 Accepted Submission(s) : 18
Problem 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
Source
PKU
题意：
给出两个四位的素数，要求求出从其中一个变化到另一个数的最少的变化次数，每一次变化只变化四位中的一位，并且变化后的数也要是素数；
思路：
bfs，只不过是40入口的bfs，需要经过剪枝；每一次都枚举个位、十位、百位、千位的所有变化，检验室素数后加入到队列中；
AC代码：
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 5 6 using namespace std; 7 int a,b; 8 struct kf 9 { 10 int number; 11 int sgin; 12 }ks[111111]; 13 bool ksgin[10100]={false}; 14 15 16 17 bool shu(int sg)//判断sg是否是素数 18 { 19 if(sg==2||sg==3) 20 return true; 21 else if(sg<=1||sg%2==0) 22 return false; 23 else if(sg>3) 24 { 25 for(int i=3;i*i<=sg;i+=2) 26 if(sg%i==0) 27 return false; 28 return true; 29 } 30 } 31 32 33 int bfs() 34 { 35 int left,right; 36 kf s; 37 ks[left=right=0].number=a; 38 ks[right++].sgin=0; 39 ksgin[a]=false; 40 while(left<right){ 41 s=ks[left++]; 42 if(s.number==b){ 43 cout<<s.sgin<<endl; 44 return 0; 45 } 46 int ge=s.number%10; 47 int shi=(s.number/10)%10; 48 for(int i=1;i<=9;i+=2){//枚举个位 49 int y=s.number/10*10+i; 50 if(y!=s.number&&ksgin[y]&&shu(y)){ 51 ksgin[y]=false; 52 ks[right].number=y; 53 ks[right++].sgin=s.sgin+1; 54 } 55 } 56 for(int i=0;i<=9;i++){//枚举十位 57 int y=s.number/100*100+i*10+ge; 58 if(y!=s.number&&ksgin[y]&&shu(y)){ 59 ksgin[y]=false; 60 ks[right].number=y; 61 ks[right++].sgin=s.sgin+1; 62 } 63 } 64 shi*=10; 65 shi+=ge; 66 for(int i=0;i<=9;i++){//枚举百位 67 int y=s.number/1000*1000+i*100+shi; 68 if(y!=s.number&&ksgin[y]&&shu(y)){ 69 ksgin[y]=false; 70 ks[right].number=y; 71 ks[right++].sgin=s.sgin+1; 72 } 73 } 74 shi=s.number%1000; 75 for(int i=1;i<=9;i++){//千位 76 int y=i*1000+shi; 77 if(y!=s.number&&ksgin[y]&&shu(y)){ 78 ksgin[y]=false; 79 ks[right].number=y; 80 ks[right++].sgin=s.sgin+1; 81 } 82 } 83 } 84 cout<<"Impossible"<<endl; 85 return 0; 86 } 87 88 89 int main() 90 { 91 // freopen("input.txt","r",stdin); 92 int test; 93 cin>>test; 94 while(test--){ 95 memset(ksgin,true,sizeof(ksgin)); 96 cin>>a>>b; 97 bfs(); 98 } 99 }