kuangbin带你飞专题POJ - 3126 Prime Path

Posted 2023-02-20 汤姆的猫生

原题重现：

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

 

大概题意：

给定两个四位数素数，起始素数为a, 终止素数为b, 求 从素数a 变成 素数b 的最短路径

条件：每次只能够变化其中一位数字， 且变化后的素数仍然为素数， 如 1033 到 3733 。

 

思路：

素数判定 + BFS。四位数中（1000到9999）中寻找到所有的素数，并存入数组。在在a开始，从素数组中搜索下一个素数（不同位数为1）。

本以为会超时，然鹅还是过了。

另外记录一个不太懂的地方，自己设了一个vis数组来标记，

为什么后面的memset初始化不管用呢（新手疯狂挠头，最后还是for重初始化）

 

代码实现

#include <iostream>
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <math.h>
#include <queue>
#include <string.h>
using namespace std;
int total_size;
int vis[10005];
vector<int> prime;
struct node 
	int val;
	int step;
	node(int val, int step):val(val),step(step)
;

//素数判定
bool isPrime(int num) 
    if (num <= 3) 
        return num > 1;
    
    if (num % 6 != 1 && num % 6 != 5) 
        return false;
    
    int sqrt_n = (int)sqrt((double)num);
    for (int i = 5; i <= sqrt_n; i += 6) 
        if (num % i == 0 || num % (i + 2) == 0) 
            return false;
        
    
    return true;


//检查下一个是否符合要求
bool checkNext(int a, int b) 
	
	int flag = 0;
	while(1) 
		int c = a % 10;
		int d = b % 10;
		a = a / 10;
		b = b / 10;
		if(c != d) flag++; 
		if(a == 0 || b == 0) break;
	
	
	if(flag == 1) return true;  //若只有一位不同，则可
	return false;  //否则，则不可


int main() 
	
 //素数组
	for(int i = 1000; i < 9999; i++) 
		if(isPrime(i)) 
			prime.push_back(i);
		
	
	
	total_size = prime.size();
	
	int T; 
	cin >> T;
 
	while(T--) 
		int a, b, flag, step;
		cin >> a >> b;
		if(a == b) 
			cout << 0 << endl;
			continue;	
		
		
  //这里不懂为什么用memset(vis, 0, sizeof(vis)) 毫无卵用呢。。
		for(int i = 0; i < 10005; i++) 
			vis[i] = 0;
		  
		
		vis[a] = 1;
		flag = 0;
		
		queue<node> sol;
		sol.push(node(a, 0));
  
  //队列完成BFS操作
		while(sol.size()) 
			node ele = sol.front();
			sol.pop();
			for(int i = 0; i < total_size; i++) 
				if(vis[prime[i]] == 1) continue;
				if(checkNext(prime[i], ele.val))  //检查下一个素数是否满足要求
					if(prime[i] == b)    //如果为结果就直接退出
						flag = 1;
						step = ele.step + 1;
						break;
					
					sol.push(node(prime[i], ele.step + 1));  //否则，进队列
					vis[prime[i]] = 1;  //这里标记已经进过队列了
				
			
			
			if(flag == 1) 
				cout << step << endl;
				break;
			
			
		
		
	
	
	
	return 0;