Prime Path
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 21581 | Accepted: 11986 |
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
Source
思路：对每位数字替换，直到成为目标数字。
代码：
1 #include "cstdio"
2 #include "iostream"
3 #include "algorithm"
4 #include "string"
5 #include "cstring"
6 #include "queue"
7 #include "cmath"
8 #include "vector"
9 #include "map"
10 #include "stdlib.h"
11 #include "set"
12 #define mj
13 #define db double
14 #define ll long long
15 using namespace std;
16 const int N=1e4+5;
17 const int mod=1e9+7;
18 const ll inf=1e16+10;
19 bool pri[N];
20 bool u[N],v[N];
21 int c[N];//统计步数
22 void init(){
23 int i,j;
24 for(i=1000;i<=N;i++){
25 for(j=2;j<i;j++)
26 if(i%j==0){
27 pri[i]=0;
28 break;
29 }
30 if(j==i) pri[i]=1;
31 }
32 }
33 int bfs(int s,int e){
34 queue<int >q;
35 memset(v,0, sizeof(v));
36 memset(c,0, sizeof(c));
37 int tmp,a[4],ans=s;
38 q.push(s);
39 v[s]=1;
40 while(q.size()){
41 int k=q.front();
42 q.pop();
43 a[0]=k/1000,a[1]=k/100%10,a[2]=k/10%10,a[3]=k%10;
44 for(int i=0;i<4;i++){
45 tmp=a[i];
46 for(int j=0;j<10;j++){
47 if(tmp!=j){
48 a[i]=j;
49 ans=a[0]*1000+a[1]*100+a[2]*10+a[3];
50 if(pri[ans]&&!v[ans]){//为素数且未使用过
51 c[ans]=c[k]+1;
52 v[ans]=1;
53 q.push(ans);
54 }
55 if(ans==e) {
56 printf("%d\n",c[ans]);
57 return 0;
58 }
59 }
60 a[i]=tmp;
61 }
62 }
63 if(k==e) {
64 printf("%d\n",c[k]);
65 return 0;
66 }
67 }
68 printf("Impossible\n");
69 return 0;
70 }
71 int main()
72 {
73 int n;
74 int x,y;
75 scanf("%d",&n);
76 init();
77 for(int i=0;i<n;i++){
78 scanf("%d%d",&x,&y);
79 bfs(x,y);
80 }
81 return 0;
82 }