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
题意：给定两个素数n和m，要求把n变成m，每次变换时只能变一个数字，即变换后的数与变换前的数只有一个数字不同，并且要保证变换后的四位数也是素数。求最小的变换次数；如果不能完成变换，输出Impossible。
无论怎么变换，个位数字一定是奇数（个位数字为偶数肯定不是素数），这样枚举个位数字时只需枚举奇数就行；而且千位数字不能是0。所以可以用广搜，枚举各个数位上的数字，满足要求的数就加入队列，直到变换成功。因为是广搜，所以一定能保证次数最少。
注意：开始数字可以不是素数。
C++版本一
BFS
#include <iostream>
#include <stdio.h>
#include <queue>
#include <math.h>
#include <string.h>
int t,a,b;
int vis[10000];
using namespace std;
struct node{
int x[4];
int step;
}f,s,p;
bool prime(int n){
for(int i=2;i<=(int)sqrt(n);i++){
if(n%i==0)
return false;
}
return true;
}
queue <node>Q;
int bfs(){
while(!Q.empty()){
Q.pop();
}
if(!prime(b)) return -1;
if(a==b)return s.step;
Q.push(s);
while(!Q.empty()){
f=Q.front();
Q.pop();
if(f.x[0]*1000+f.x[1]*100+f.x[2]*10+f.x[3]==b)
return f.step;
for(int i=0;i<4;i++){
for(int j=0;j<=9;j++){
if(f.x[i]!=j){
p=f;
p.x[i]=j;//printf("%d\n",p.x[0]*1000+p.x[1]*100+p.x[2]*10+p.x[3]);
int temp=p.x[0]*1000+p.x[1]*100+p.x[2]*10+p.x[3];
if(vis[temp]==0&&p.x[0]!=0&&prime(temp)){
p.step++;
vis[temp]=1;
Q.push(p);
}
}
}
}
}
return -1;
}
int main()
{
scanf("%d",&t);
while(t--){
scanf("%d%d",&a,&b);
memset(vis,0,sizeof(vis));
s.x[0]=a/1000;
s.x[1]=a%1000/100;
s.x[2]=a%100/10;
s.x[3]=a%10;
s.step=0;
vis[a]=1;
int ans=bfs();
if(ans>=0)printf("%d\n",ans);
else printf("Impossible\n");
}
//cout << "Hello world!" << endl;
return 0;
}
C++版本二
BFS
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
int n, m;
const int N = 1e4 + 100;
int vis[N];
struct node
{
int x, step;
};
queue<node> Q;
bool judge_prime(int x) //判断素数
{
if(x == 0 || x == 1)
return false;
else if(x == 2 || x == 3)
return true;
else
{
for(int i = 2; i <= (int)sqrt(x); i++)
if(x % i == 0)
return false;
return true;
}
}
void BFS()
{
int X, STEP, i;
while(!Q.empty())
{
node tmp;
tmp = Q.front();
Q.pop();
X = tmp.x;
STEP = tmp.step;
if(X == m)
{
printf("%d\n",STEP);
return ;
}
for(i = 1; i <= 9; i += 2) //个位
{
int s = X / 10 * 10 + i;
if(s != X && !vis[s] && judge_prime(s))
{
vis[s] = 1;
node temp;
temp.x = s;
temp.step = STEP + 1;
Q.push(temp);
}
}
for(i = 0; i <= 9; i++) //十位
{
int s = X / 100 * 100 + i * 10 + X % 10;
if(s != X && !vis[s] && judge_prime(s))
{
vis[s] = 1;
node temp;
temp.x = s;
temp.step = STEP + 1;
Q.push(temp);
}
}
for(i = 0; i <= 9; i++) //百位
{
int s = X / 1000 * 1000 + i * 100 + X % 100;
if(s != X && !vis[s] && judge_prime(s))
{
vis[s] = 1;
node temp;
temp.x = s;
temp.step = STEP + 1;
Q.push(temp);
}
}
for(i = 1; i <= 9; i++) //千位
{
int s = i * 1000 + X % 1000;
if(s != X && !vis[s] && judge_prime(s))
{
vis[s] = 1;
node temp;
temp.x = s;
temp.step = STEP + 1;
Q.push(temp);
}
}
}
printf("Impossible\n");
return ;
}
int main()
{
int t, i;
scanf("%d",&t);
while(t--)
{
while(!Q.empty()) Q.pop();
scanf("%d%d",&n,&m);
memset(vis,0,sizeof(vis));
vis[n] = 1;
node tmp;
tmp.x = n;
tmp.step = 0;
Q.push(tmp);
BFS();
}
return 0;
}
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