POJ 3126, Prime Path

POJ 3126, Prime Path

Time Limit: 1000MS  Memory Limit: 65536K
Total Submissions: 3134  Accepted: 1911

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

 这个题目是比较好的题目，看起来是数学题其实是个搜索题，而且每个搜索的状态转移也比较新颖，是把一个四位的数字的每一位换成是0-9；

代码：

View Code

 1 #include <iostream>
 2 #include <queue>
 3 #include <cstdio>
 4 #include <cstring>
 5 
 6 using namespace std;
 7 
 8 int main()
 9 {
10     //init prime table
11     bool prime[10001];
12     memset(prime,1,sizeof (prime));
13     prime[0] = false;prime[1] = false;
14     for (int i = 2; i < 5000; ++i)
15     {
16         int k = 2;
17         int next;
18         while((next = (k++) * i) < 10000)
19             prime[next] = false;
20     }
21 
22     int cases;
23     cin >> cases;
24     int beg, end;
25     int step;
26     int visited[10001];
27     int base[4]= {1, 10, 100, 1000};
28     for (int c = 0; c < cases; ++c)
29     {
30         scanf("%d %d", &beg, &end);
31         memset(visited, -1, sizeof(visited));
32         queue<int> q;
33         q.push(beg);
34         visited[beg] = 0;
35         step = -1;
36 
37         while (!q.empty())
38         {
39             int cur = q.front();
40             q.pop();
41             if (cur == end)
42             {
43                 step = visited[cur];
44                 break;
45             }
46             
47             for (int i = 0; i < 4; ++i)
48             {
49                 for (int j = 0; j < 10; ++j)
50                 {
51                     int l = cur / (base[i] * 10);
52                     int r = cur % base[i];//base[i]数组很巧妙很好。值得学习
53                     int num = l * base[i] * 10 + j * base[i] + r;
54                     if (num > 1000 && num != cur && prime[num] == true && visited[num] == -1)
55                     {
56                         q.push(num);
57                         visited[num] = visited[cur] + 1;
58                     }
59                 }
60             }
61         }
62 
63         if (step == -1)cout << "Impossible\n";
64         else cout << step << endl;
65     }
66 
67     return 0;
68 }

转载于:https://www.cnblogs.com/one--world--one--dream/archive/2011/10/29/2228880.html