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 题目大意：n组输入，每一组输入两个四位质数，对第一个质数进行转换，使之变为第二个质数，每次只能更改一个数中某一位的数，且变动一位后的数也是质数才符合要求。输出最少经过几次转换，能转换成第二个质数，保证输入合法。 思路：我这里采用了比较笨的方法，用结构体存储每一位上的数，以及总数，建立结构体队列，进行广度优先搜索，通过数组存储转换次数信息。（主要是自己比较菜，获取某个数中每一位的数不够熟练，后续会补上用整型解决问题的思路。）还有一个问题是判断某数是否为质数的函数，这里我参考了大佬的博客（https://blog.csdn.net/huang_miao_xin/article/details/51331710），很巧妙的判断质数的算法，值得学习。代码如下：
#include <iostream>
#include <algorithm>
#include <math.h>
#include <queue>
#include <string.h>
int const max_n = 10002;
using namespace std;
int jug[max_n];//记录信息数组
struct node {
int a, b, c, d;//千位，百位，十位，个位
int sum;//数的大小
};
queue<node> q;
int test(node x)//由各个数位的数求大小
{
return x.a * 1000 + x.b * 100 + x.c * 10 + x.d;
}
bool judge(int x)//是否为质数
{//关于这种方法判断质数的解释，6x，6x+2，6x+4一定被2整除，6x+3一定被3整除，这样就只用判断6x+1和6x+5的数了。而当循环以6位单位跨进时，就是6x-1和6x+1的情况了。更详细的讲解可以去大佬博客看
if (x == 2 || x == 3)return true;//特判2,3两个质数
if (x % 6 != 1 && x % 6 != 5)return false;
int tmp = sqrt(x);
for (int i = 5; i <= tmp; i += 6)
{
if (x%i == 0 || x % (i + 2) == 0)return false;
}
return true;
}
int bfs(int x, int y)
{
node sx;
sx.a = x / 1000;
sx.b = x / 100 % 10;
sx.c = x / 10 % 10 % 10;
sx.d = x % 1000 % 100 % 10;
sx.sum = x;
q.push(sx);//入队
memset(jug, 0, sizeof(jug));
jug[x] = 1;
while (!q.empty())
{
node s = q.front(); q.pop();
if (s.sum == y)return jug[s.sum] - 1;
for (int i = 1; i < 10; i++)//千位进行转换
{
node p;
p.a = s.a, p.b = s.b, p.c = s.c, p.d = s.d;
p.a = i;
p.sum = test(p);
if (judge(p.sum) && jug[p.sum] == 0)
{
jug[p.sum] = jug[s.sum] + 1;
q.push(p);
}
}
for (int i = 0; i < 10; i++)//百位进行转换
{
node p;
p.a = s.a, p.b = s.b, p.c = s.c, p.d = s.d;
p.b = i;
p.sum = test(p);
if (judge(p.sum) && jug[p.sum] == 0)
{
jug[p.sum] = jug[s.sum] + 1;
q.push(p);
}
}
for (int i = 0; i < 10; i++)//十位进行转换
{
node p;
p.a = s.a, p.b = s.b, p.c = s.c, p.d = s.d;
p.c = i;
p.sum = test(p);
if (judge(p.sum) && jug[p.sum] == 0)
{
jug[p.sum] = jug[s.sum] + 1;
q.push(p);
}
}
for (int i = 0; i < 10; i++)//个位进行转换
{
node p;
p.a = s.a, p.b = s.b, p.c = s.c, p.d = s.d;
p.d = i;
p.sum = test(p);
if (judge(p.sum) && jug[p.sum] == 0)
{
jug[p.sum] = jug[s.sum] + 1;
q.push(p);
}
}
}
return 0;
}
int main()
{
int t; scanf("%d", &t);
while (t--)
{
int m, n;
scanf("%d %d", &m, &n);
while (q.size())q.pop();
if (m == n) { printf("0\n"); continue; }
else
{
int x = bfs(m, n);
if (x == 0)printf("Impossible\n");
else printf("%d\n", x);
}
}
return 0;
}
原文地址：https://www.cnblogs.com/whocarethat/p/11099777.html