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.
Input
— 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
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
总结：
该题目的大体意思是：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。因为不是求全部（全部的话用深搜），只是让求的变换后的数字与变化前的只有一位不相同，用广度搜索最为合适。
注意：
1.bfs()最后返回的值，应当注意一下。例如题目中的第三个样例，是两个相同的数据输入时，输出的结果是0,那么此时就不可以将0作为函数的返回值。
2.每一个新加入到队列中的vtemp，就是下一次while(!q.empty())里面新的v，count[v]的变化其实就是上一次的count[vtemp]。（只要队列不空while()循环一直进行，以1033 3733为例，它的while()循环就进行了1941次）
问题：
1.下面代码是通过的，可是存在一个问题，加入将素数不进行打表而是进行单纯的判断，输出的结果就会存在问题。（将以下非注释部分换成注释部分就有问题，还请大家指教）
/*int isPrime(int m)
{
int j;
for(j=2;j<=m/2;j++)
{
if(m%j==0)
{
return 0;
}
}
if(j>m/2) return 1;
} */
void init()
{
int j,i;
for(i=1000;i<10000;i++)
{
for(j=2;j<i;j++)
{
if(i%j==0)
{
vis[i]=false;
break;
}
}
if(j==i) vis[i]=true;
}
}
做过单独的验证，真的没有发现这两个方法存在什么不同（可气）：
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
int num1[10005];
int num2[10005];
int f1=0;
int f2=0;
void init()
{
int i,j;
for(i=1000;i<=10000;i++)
{
for(j=2;j<i;j++)
{
if(i%j==0)
{
break;
}
}
if(i==j) num1[f1++]=i;
}
}
void isprime()
{
int i,j;
for(i=1000;i<=10000;i++)
{
for(j=2;j<=sqrt(i);j++)
{
if(i%j==0)
{
break;
}
}
if(j>sqrt(i)) num2[f2++]=i;
}
}
int main()
{
init();
isprime();
cout<<"f1="<<f1<<endl;
cout<<"f2="<<f2<<endl;
/*for(int i=0;i<=f1;i++)
cout<<"1:"<<num1[i]<<endl;
for(int j=0;j<f2;j++)
cout<<"2:"<<num2[j]<<endl;*/
return 0;
}#include <iostream>
#include <queue>
#include<cstring>
#include<cmath>
#include<cstdio>
using namespace std;
bool vis[10005];
/*int isPrime(int m)
{
int j;
for(j=2;j<=m/2;j++)
{
if(m%j==0)
{
return 0;
}
}
if(j>m/2) return 1;
} */
void init()
{
int j,i;
for(i=1000;i<10000;i++)
{
for(j=2;j<i;j++)
{
if(i%j==0)
{
vis[i]=false;
break;
}
}
if(j==i) vis[i]=true;
}
}
int bfs(int first,int last)
{
bool dis[10005];
queue <int>q;
int v,i,j,temp,vtemp,count[10005],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/10)%10;
t[3]=v%10;
for(j=0;j<4;j++)
{
temp=t[j];//下面会有变化，一定得保存
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] &&vis[vtemp])//isPrime(vtemp))
{
count[vtemp]=count[v]+1;
dis[vtemp]=true;
//cout<<"j="<<j<<",i="<<i<<",vtemp="<<vtemp<<",count[vtemp]="<<count[vtemp]<<",v="<<v<<",count[v]="<<count[v]<<endl;
q.push(vtemp);
}
if(vtemp==last) return count[vtemp];
}
t[j]=temp;
}
if(v==last) return count[v];
}
return -1;
}
int main()
{
int n,num1,num2;
init();
cin>>n;
while(n--)
{
cin>>num1>>num2;
if(bfs(num1,num2)==-1)
cout<<"Impossible"<<endl;
else
cout<<bfs(num1,num2)<<endl;
}
return 0;
}