Prime Path（BFS）

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.

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

1033 8179

1373 8017

1033 1033

Sample Output

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 }

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