转载请注明出处：https://blog.csdn.net/Mercury_Lc/article/details/82697622 作者：Mercury_Lc
题意：就是给你一个n，让你每次可以改变n的位数上的一个数，每次操作完必须是素数，要求最小次数的改变到达m。
题解：对n每一位都进行判断，找到通过最小操作次数得到m。分别要从个位、十位、百位、千位判断，在个位的时候每次只能是1、3、5、7、9，其他的改变之后都不是素数，十位、百位、千位都从0开始遍历到9，每次只要符合是素数就放到队列中，开个结构体记录步数和当前的数就可以了。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <iostream>
using namespace std;
const int maxn = 1e6;
int n,m;
int vis[maxn];
struct node
{
int data,step;
} w,l;
bool prime(int x)
{
if(x==1||x==0)
return 0;
for(int i = 2; i <= sqrt(x); i ++)
{
if(x % i == 0)
return 0;
}
return 1;
}
void bfs()
{
queue<node>q;
memset(vis,0,sizeof(vis));
vis[n] = 1;
w.data = n;
w.step = 0;
q.push(w);
while(!q.empty())
{
w = q.front();
q.pop();
if(w.data == m)
{
printf("%d\n",w.step);
return ;
}
for(int i = 1; i <= 9; i += 2) // ge
{
int s = w.data / 10 * 10 + i;
if(!vis[s] && prime(s))
{
vis[s] = 1;
l.data = s;
l.step = w.step + 1;
q.push(l);
}
}
for(int i = 0; i <= 9; i++) // shi
{
int s = w.data / 100 * 100 + i * 10 + w.data % 10;
if(!vis[s] && prime(s))
{
vis[s] = 1;
l.data = s;
l.step = w.step + 1;
q.push(l);
}
}
for(int i = 0; i <= 9; i++) // bai
{
int s = w.data / 1000 * 1000 + i * 100 + w.data % 100;
if(!vis[s] && prime(s))
{
vis[s] = 1;
l.data = s;
l.step = w.step + 1;
q.push(l);
}
}
for(int i = 1; i <= 9; i++) // qian
{
int s = i * 1000 + w.data % 1000;
if(!vis[s] && prime(s))
{
vis[s] = 1;
l.data = s;
l.step = w.step + 1;
q.push(l);
}
}
}
return ;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
bfs();
}
return 0;
}
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.
1033
— 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.
1733
3733
3739
3779
8779
8179The 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 1033Sample Output
6
7
0