題目連結：https://vjudge.net/problem/POJ-3126

從一個素數到另一個素數，每一步只能改變一個數，途徑的數也必須是素數，求最小步數。

AC程式碼：

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstring>
 4 #include<cstdio>
 5 #include<cmath>
 6 #include<stack>
 7 #include<queue>
 8 
 9 using namespace std;
10
 
 typedef long long ll;
11 const int maxn = 2e6 + 5;
12 const int INF = 0x3f3f3f3f;
13 
14 bool isPrime[10005];
15 int vis[10005];
16 int prime[10005];
17 
18 void Prime(int R)
19 {
20     memset(isPrime, 0, sizeof(isPrime));
21     memset(prime, 0, sizeof(prime));
22     for (int i = 2; i <= R; i++) {
23         if
 
 (!isPrime[i]) {
24             prime[++prime[0]] = i;
25         }
26         for (int j = 1; j <= prime[0] && i * prime[j] <= R; j++) {
27             isPrime[i * prime[j]] = 1;
28             if (i % prime[j] == 0) break;
29         }
30     }
31 }
32 
33 struct node
34 {
35     int num;
 
36     int step;
37 };
38 
39 int st, ed;
40 
41 int bfs(int st)
42 {
43     queue<node>qe;
44     node start, next;
45     start.num = st;
46     start.step = 0;
47     memset(vis, 0, sizeof(vis));
48     qe.push(start);
49     while (!qe.empty()) {
50         node temp = qe.front();
51         qe.pop();
52         if (temp.num == ed) {
53             return temp.step;
54         }
55         vis[temp.num] = 1;
56         for (int k = 0; k < 4; k++) {
57             int head = 0;
58             int x = (int)pow(10.0, k);
59             head = temp.num - (temp.num / x % 10) * x;
60             for (int i = 0; i <= 9; i++) {
61                 if (x == 1000 && i == 0) continue;
62                 next.num = head + i * x;
63                 if (next.num > 9999) break;
64                 if (!isPrime[next.num] && !vis[next.num]) {
65                     next.step = temp.step + 1;
66                     qe.push(next);
67                     vis[next.num] = 1;
68                 }
69             }
70         }
71     }
72     return -1;
73 }
74 
75 int main()
76 {
77     Prime(10001);
78     int t;
79     cin >> t;
80     while (t--)
81     {
82         cin >> st >> ed;
83         int ans = bfs(st);
84         cout << ans << endl;
85     }
86     return 0;
87 }