[HDU 1973]--Prime Path(BFS,素数表)

 1 #include <iostream>
 2 #include <queue>
 3 #include <cstdio>
 4 #include <cstring>
 5 using namespace std;
 6 struct node{
 7     int cur, step;
 8 }now, Next;
 9 int vis[10001], star, finish, prime[10001] = { 0, 0, 1 };
10 void init(){
11     for (int i = 2; i < 10001; i++){
12         if (!prime[i]){
13             for (int j = 2; i*j < 10001; j++)
14                 prime[i*j] = 1;
15         }
16     }
17 }
18 int bfs(){
19     queue<node> Q;
20     vis[star] = 1;
21     now.cur = star, now.step = 0;
22     Q.push(now);
23     while (!Q.empty()){
24         int i, j;
25         char num[5];
26         now = Q.front();
27         Q.pop();
28         if (now.cur == finish) return now.step;
29         for (i = 0; i < 4; i++){
30             sprintf(num, "%d", now.cur);
31             for (j = 0; j < 10; j++){
32                 if (j == 0 && i == 0)
33                     continue;
34                 if (i == 0)
35                     Next.cur = j * 1000 + (num[1] - ‘0‘) * 100 + (num[2] - ‘0‘) * 10 + (num[3] - ‘0‘);
36                 else if (i == 1)
37                     Next.cur = j * 100 + (num[0] - ‘0‘) * 1000 + (num[2] - ‘0‘) * 10 + (num[3] - ‘0‘);
38                 else if (i == 2)
39                     Next.cur = j * 10 + (num[0] - ‘0‘) * 1000 + (num[1] - ‘0‘) * 100 + (num[3] - ‘0‘);
40                 else if (i == 3)
41                     Next.cur = j + (num[0] - ‘0‘) * 1000 + (num[1] - ‘0‘) * 100 + (num[2] - ‘0‘) * 10;
42                 if (!prime[Next.cur] && !vis[Next.cur])
43                 {
44                     Next.step = now.step + 1;
45                     vis[Next.cur] = 1;
46                     Q.push(Next);
47                 }
48             }
49         }
50     }
51     return -1;
52 }
53 int main(){
54     int t, ans;
55     cin >> t;
56     init();
57     while (t--){
58         cin >> star >> finish;
59         memset(vis, 0, sizeof(vis));
60         ans = bfs();
61         if (ans == -1) cout << "Impossible\n";
62         else cout << ans << endl;
63     }
64     return 0;
65 }