• 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

    Source
    Northwestern Europe 2006


    // POJ3126.cpp : Defines the entry point for the console application.
    //

    #include 
    <iostream>
    #include 
    <queue>
    using namespace std;

    int main(int argc, char* argv[])
    {
        
    //init prime table
        bool prime[10001];
        memset(prime,
    1,sizeof (prime));
        prime[
    0= false;prime[1= false;
        
    for (int i = 2; i < 5000++i)
        {
            
    int k = 2;
            
    int next;
            
    while((next = (k++* i) < 10000)
                prime[next] 
    = false;
        }

        
    int cases;
        cin 
    >> cases;
        
    int beg, end;
        
    int step;
        
    int visited[10001];
        
    int base[4]= {1101001000};
        
    for (int c = 0; c < cases; ++c)
        {
            scanf(
    "%d %d"&beg, &end);
            memset(visited, 
    -1sizeof(visited));
            queue
    <int> q;
            q.push(beg);
            visited[beg] 
    = 0;
            step 
    = -1;

            
    while (!q.empty())
            {
                
    int cur = q.front();
                q.pop();
                
    if (cur == end)
                {
                    step 
    = visited[cur];
                    
    break;
                }
                
                
    for (int i = 0; i < 4++i)
                {
                    
    for (int j = 0; j < 10++j)
                    {
                        
    int l = cur / (base[i] * 10);
                        
    int r = cur % base[i];
                        
    int num = l * base[i] * 10 + j * base[i] + r;
                        
    if (num > 1000 && num != cur && prime[num] == true && visited[num] == -1)
                        {
                            q.push(num);
                            visited[num] 
    = visited[cur] + 1;
                        }
                    }
                }
            }

            
    if (step == -1)cout << "Impossible\n";
            
    else cout << step << endl;
        }

        
    return 0;
    }

  • 相关阅读:
    linux内核编译
    字符设备驱动ioctl实现用户层内核层通信
    Linux内核完全剖析基于0.12内核
    KVM分析报告
    kvm的vmcall
    kvm源代码分析
    KVM基本概念
    linux系统调用
    UML的9种图例解析(转)
    SurfaceView的基本使用(转)
  • 原文地址:https://www.cnblogs.com/asuran/p/1580732.html
Copyright © 2020-2023  润新知