Prime Path (POJ - 3126)
题面
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.
输入
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).
输出
One line for each case, either with a number stating the minimal cost or containing the word Impossible.
样例输入
13
21033 8179
31373 8017
41033 1033
样例输出
16
27
30
提示
无
思路
欧拉筛打表，枚举每一位的变化，注意大于2的偶数不是质数
代码
1using namespace std;
2typedef long long ll;
3const int inf = 0x3f3f3f3f;
4const int N = 1e5+5;
5
6bool isp[N] = {0};
7bool vis[N] = {0};
8int n, s, e;
9
10struct P {
11 int x;
12 int step;
13};
14
15void euler() {
16 int p[N], m=0;
17 for(int i=2; i<=N; i++) {
18 if(!isp[i])
19 p[m++] = i;
20 for(int j=0; j<m; j++) {
21 if(p[j]*i>N)
22 break;
23 isp[p[j]*i] = 1;
24 if(i%p[j]==0)
25 break;
26 }
27 }
28}
29
30int bfs() {
31 memset(vis, 0, sizeof(vis));
32
33 P sp;
34 sp.x = s;
35 sp.step = 0;
36
37 queue<P> q;
38 q.push(sp);
39 vis[sp.x] = 1;
40
41 while(!q.empty()) {
42 P tp = q.front();
43 q.pop();
44
45 if(tp.x==e) {
46 return tp.step;
47 }
48 P np = tp;
49 np.step = tp.step+1;
50
51 for(int i=1; i<=9; i+=2) {
52 np.x = tp.x/10*10 + i;
53 if(np.x!=tp.x && !vis[np.x] && !isp[np.x]) {
54 q.push(np);
55 vis[np.x] = 1;
56 }
57 }
58 for(int i=0; i<=9; i++) {
59 np.x = tp.x/100*100 + i*10 + tp.x%10;
60 if(np.x!=tp.x && !vis[np.x] && !isp[np.x]) {
61 q.push(np);
62 vis[np.x] = 1;
63 }
64 }
65 for(int i=0; i<=9; i++) {
66 np.x = tp.x/1000*1000 + i*100 + tp.x%100;
67 if(np.x!=tp.x && !vis[np.x] && !isp[np.x]) {
68 q.push(np);
69 vis[np.x] = 1;
70 }
71 }
72 for(int i=1; i<=9; i++) {
73 np.x = i*1000 + tp.x%1000;
74 if(np.x!=tp.x && !vis[np.x] && !isp[np.x]) {
75 q.push(np);
76 vis[np.x] = 1;
77 }
78 }
79 }
80 return 0;
81}
82
83int main(void) {
84 euler();
85 scanf("%d", &n);
86 for(int i=0; i<n; i++) {
87 scanf("%d%d", &s, &e);
88 printf("%d\n", bfs());
89 }
90 return 0;
91}