链接：戳这里
Prime Path
Time Limit: 1000MS
Memory Limit: 65536K
Description
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：
两个1000~9999的素数x,y,要求在改变千、百、十、个位上的数字得到一个新的素数
问最少经过几次变换可以使得x->y 注意全程的数都必须是素数
思路：
处理一下1000~9999的素数,并标记
然后暴力改成新的数再去判断 bfs
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
int mu[1000100],prime[1000100],vis[1000100],a[1000100];
struct node{
int v,step;
node(int v=0,int step=0):v(v),step(step){}
};
int used[1000100];
void BFS(int x,int y){
mst(used,0);
queue<node> qu;
qu.push(node(x,0));
used[x]=1;
while(!qu.empty()){
node now=qu.front();
qu.pop();
if(now.v==y){
cout<<now.step<<endl;
return ;
}
for(int i=0;i<4;i++){
for(int j=0;j<=9;j++){
if(i==0 && j==0) continue;
if(i==0){
int tmp=now.v%1000+j*1000;
///printf("%d %d %d %d\n",i,j,now.v,tmp);
if(!used[tmp] && vis[tmp]){
used[tmp]=1;
qu.push(node(tmp,now.step+1));
}
} else if(i==1){
int tmp=now.v%100+j*100+(now.v/1000)*1000;
if(!used[tmp] && vis[tmp]){
used[tmp]=1;
qu.push(node(tmp,now.step+1));
}
} else if(i==2){
int tmp=now.v/100*100+j*10+now.v%10;
if(!used[tmp] && vis[tmp]){
used[tmp]=1;
qu.push(node(tmp,now.step+1));
}
} else {
int tmp=now.v/10*10+j;
if(!used[tmp] && vis[tmp]){
used[tmp]=1;
qu.push(node(tmp,now.step+1));
}
}
}
}
}
}
int main(){
mu[1]=1;
int cnt=0,num=0;
for(int i=2;i<=100000;i++){
if(!vis[i]){
prime[cnt++]=i;
mu[i]=-1;
if(i>=1000 && i<10000) a[num++]=i;
}
for(int j=0;j<cnt;j++){
if(i*prime[j]>100000) break;
vis[i*prime[j]]=1;
if(i%prime[j]==0){
mu[i*prime[j]]=0;
break;
} else mu[i*prime[j]]=-mu[i];
}
}
mst(vis,0);
for(int i=0;i<num;i++) vis[a[i]]=1;
int T;
scanf("%d",&T);
while(T--){
int x,y;
scanf("%d%d",&x,&y);
BFS(x,y);
}
return 0;
}