poj3126 Prime Path

题意：

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

思路：

bfs

实现：

1 #include <iostream>
2 #include <cstdio>
3 #include <queue>
4 #include <cstring>
5 using namespace std;
6
7 const int MAXN = 10000;
8 bool check[MAXN + 5];
9 bool vis[MAXN + 5];
10
11 struct node
12 {
13     int now, d;
14 };
15
16 void init()
17 {
18     for (int i = 0; i <= MAXN; i++)
19     {
20         check[i] = true;
21     }
22     check[0] = check[1] = false;
23     for (int i = 2; i <= MAXN; i++)
24     {
25         if (check[i])
26         {
27             for (int j = 2 * i; j <= MAXN; j += i)
28                 check[j] = false;
29         }
30     }
31 }
32
33 bool is_prime(int x)
34 {
35     return check[x];
36 }
37
38 int solve(int x, int t)
39 {
40     node start;
41     start.now = x;
42     start.d = 0;
43     vis[x] = true;
44     queue<node> q;
45     q.push(start);
46     while (!q.empty())
47     {
48         node tmp = q.front();
49         q.pop();
50         if (tmp.now == t)
51         {
52             return tmp.d;
53         }
54         int u = tmp.now;
55         int t = 1000;
56         int last = 0;
57         for (int i = 0; i < 4; i++)
58         {
59             int now = u / t % 10;
60             for (int j = 0; j <= 9; j++)
61             {
62                 if ((i == 0 && j == 0) || j == now)
63                     continue;
64                 int n = last + j * t + u % t;
65                 if (is_prime(n) && !vis
)
66                 {
67                     vis
= true;
68                     node son;
69                     son.now = n;
70                     son.d = tmp.d + 1;
71                     q.push(son);
72                 }
73             }
74             last += now * t;
75             t /= 10;
76         }
77     }
78     return -1;
79 }
80
81 int main()
82 {
83     int t, x, y;
84     cin >> t;
85     init();
86     while (t--)
87     {
88         cin >> x >> y;
89         memset(vis, 0, sizeof(vis));
90         int res = solve(x, y);
91         if (res != -1)
92             cout << res << endl;
93         else
94             cout << "Impossible" << endl;
95     }
96     return 0;
97 }