Prime Path
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 26158 | Accepted: 14376 |
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.
— 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.
1033The 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.
1733
3733
3739
3779
8779
8179
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
Source
题意：
给定两个素数a b，求a变到b需要几步，并且变换时只有一个数字不同，并且是素数
思路：
因为是四位的素数，所以先对素数打表，然后BFS，每次进入队列的是：只有一个数字不同的四位数，并且是素数，就入队
#include<iostream>
#include<cstring>
#include<queue>
#include<cmath>
#include<cstdio>
using namespace std;
#define MAXV 10009
bool prime[MAXV];
void init(){ //对素数打表
int i,j;
for(i=1000;i<=MAXV;i++){
for(j=2;j<i;j++)
if(i%j==0){
prime[i]=false;
break;
}
if(j==i) prime[i]=true;
}
/*
memset(prime,0,sizeof(prime));
int m=(int)sqrt(MAXV+0.5);
for(int i=2;i<=m;i++)
if(!prime[i])
for(int j=i*i;j<MAXV;j+=i)
prime[j]=1;*/
}
int bfs(int first,int last){
bool dis[MAXV];
queue <int>q;
int v,i,j,temp,vtemp,count[MAXV],t[4];
memset(dis,false,sizeof(dis));
memset(count,0,sizeof(count));
q.push(first);
dis[first]=true;
while(!q.empty()){
v=q.front();
q.pop();
t[0]=v/1000;
t[1]=v%1000/100;
t[2]=v%100/10;
t[3]=v%10;
// printf("%d %d %d %d",t[0],t[1],t[2],t[3]);
for(j=0;j<4;j++){
temp=t[j];
for(i=0;i<10;i++)
if(i!=temp){
t[j]=i;
vtemp=t[0]*1000+t[1]*100+t[2]*10+t[3];
if(!dis[vtemp] && prime[vtemp]){
count[vtemp]=count[v]+1;
dis[vtemp]=true;
q.push(vtemp);
}
if(vtemp==last) return count[vtemp];
}
t[j]=temp;
}
if(v==last) return count[v];
}
return -1;
}
int main(){
int n,a,b,key;
init();
scanf("%d",&n);
while(n--){
scanf("%d%d",&a,&b);
key=bfs(a,b);
if(key!=-1) printf("%d\n",key);
else printf("Impossible\n");
}
return 0;
}