首页 > 专题系列 > Java解POJ > POJ 3126 Prime Path [解题报告] Java
2013
11-12

POJ 3126 Prime Path [解题报告] Java

Prime Path

问题描述 :

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.

输入:

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).

输出:

One line for each case, either with a number stating the minimal cost or containing the word Impossible.

样例输入:

3
1033 8179
1373 8017
1033 1033

样例输出:

6
7
0

解题代码:

import java.io.BufferedInputStream;   
import java.util.LinkedList;   
import java.util.Scanner;   
public class Main {   
  
    static boolean[] isPrime = Prime.getPrimes(10000);   
  
    public static void main(String[] args) {   
  
        Scanner scan = new Scanner(new BufferedInputStream(System.in));   
        int cas = scan.nextInt();   
        for (int i = 1; i <= cas; i++) {   
            int start = scan.nextInt();   
            int end = scan.nextInt();   
            boolean[] isVisited = new boolean[10000];   
            int[] step = new int[10000];   
            LinkedList queue = new LinkedList();   
            queue.addLast(start);   
            isVisited[start] = true;   
            while (!queue.isEmpty()) {   
                int current = queue.pop();   
                if (current == end) {   
                    break;   
                }   
                for (int j = 0; j <= 9; j++) {   
                    int next1 = getNext(1, j, current);   
                    int next2 = getNext(2, j, current);   
                    int next3 = getNext(3, j, current);   
                    int next4 = getNext(4, j, current);   
                    if (!isVisited[next1]) {   
                        queue.addLast(next1);   
                        step[next1] = step[current] + 1;   
                        isVisited[next1] = true;   
                    }   
                    if (!isVisited[next2]) {   
                        queue.addLast(next2);   
                        step[next2] = step[current] + 1;   
                        isVisited[next2] = true;   
                    }   
                    if (!isVisited[next3]) {   
                        queue.addLast(next3);   
                        step[next3] = step[current] + 1;   
                        isVisited[next3] = true;   
                    }   
                    if (!isVisited[next4]) {   
                        queue.addLast(next4);   
                        step[next4] = step[current] + 1;   
                        isVisited[next4] = true;   
                    }   
                }   
            }   
            System.out.println(step[end]);   
        }   
    }   
  
    public static int getNext(int flag, int i, int current) {   
        int next = 0;   
        if (flag == 1) {   
            if (i == 0) {   
                return current;   
            }   
            next = current % 1000 + i * 1000;   
        }   
        if (flag == 2) {   
            int t = current / 1000;   
            next = t * 1000 + current % 1000 % 100 + i * 100;   
        }   
        if (flag == 3) {   
            int t = current / 100;   
            int tt = current % 10;   
            next = t * 100 + i * 10 + tt;   
        }   
        if (flag == 4) {   
            next = current / 10 * 10 + i;   
  
        }   
        if (!isPrime[next]) {   
            return current;   
        }   
        return next;   
    }   
}   
  
class Prime {   
  
    public static boolean[] getPrimes(int n) {   
        int i, j, k, x;   
        boolean[] a = new boolean[n];   
        n++;   
        n /= 2;   
        int[] b = new int[(n + 1) * 2];   
        a[2] = true;   
        a[3] = true;   
        for (i = 1; i <= 2 * n; i++) {   
            b[i] = 0;   
        }   
        for (i = 3; i <= n; i += 3) {   
            for (j = 0; j < 2; j++) {   
                x = 2 * (i + j) - 1;   
                while (b[x] == 0) {   
                    a[x] = true;   
                    for (k = x; k <= 2 * n; k += x) {   
                        b[k] = 1;   
                    }   
                }   
            }   
        }   
        return a;   
    }   
}

  1. 我没看懂题目
    2
    5 6 -1 5 4 -7
    7 0 6 -1 1 -6 7 -5
    我觉得第一个应该是5 6 -1 5 4 输出是19 5 4
    第二个是7 0 6 -1 1 -6 7输出是14 7 7
    不知道题目例子是怎么得出来的