• POJ3126(KB1-F BFS)


    Prime Path

     

    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

    Source

     
     1 //2017-02-23
     2 #include <iostream>
     3 #include <cstdio>
     4 #include <cstring>
     5 #include <queue>
     6 
     7 using namespace std;
     8 
     9 struct node
    10 {
    11     int num, step;
    12 };
    13 
    14 bool isPrime(int n)
    15 {
    16     for(int i = 2; i*i <= n; i++)
    17           if(n%i==0)return false;
    18     return true;
    19 }
    20 
    21 int pow(int a, int n)
    22 {
    23     int ans = 1;
    24     for(int i = 0; i < n; i++)
    25         ans *= a;
    26     return ans;
    27 }
    28 
    29 int main()
    30 {
    31     int n, x, y;
    32     bool book[10005];
    33     cin>>n;
    34     while(n--)
    35     {
    36         cin>>x>>y;
    37         queue<node> q;
    38         node tmp;
    39         tmp.num = x;
    40         tmp.step = 0;
    41         q.push(tmp);
    42         int num, step;
    43         bool ok = false;
    44         memset(book, 0, sizeof(book));
    45         book[x] = 1;
    46         while(!q.empty()){
    47             num = q.front().num;
    48             step = q.front().step;
    49             q.pop();
    50             if(num == y){
    51                 ok = true;
    52                 cout<<step<<endl;
    53                 break;
    54             }
    55             for(int i = 0; i < 4; i++){
    56                 for(int p = 0; p < 10; p++){
    57                     if(i==3 && p==0)continue;
    58                     int nowNum = num-((num/pow(10, i))%10)*pow(10, i)+p*pow(10, i);
    59                     if(!book[nowNum] && isPrime(nowNum)){
    60                         tmp.num = nowNum;
    61                         tmp.step = step+1;
    62                         q.push(tmp);
    63                         book[nowNum] = 1;
    64                     }
    65                 }
    66             }
    67             if(ok)break;
    68         }
    69     }
    70 
    71     return 0;
    72 }
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  • 原文地址:https://www.cnblogs.com/Penn000/p/6435409.html
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