Prime Path
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 28470 | Accepted: 15534 |
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 1033Sample Output
6
7
0Source
问题链接：POJ3126 Prime Path
问题描述：给定两个四位素数a b，要求把a变换到b。变换的过程要保证每次变换出来的数都是一个 四位素数，而且当前这步的变换所得的素数 与 前一步得到的素数只能有一个位不同，而且每步得到的素数都不能重复。求从a到b最少需要的变换次数。无法变换则输出Impossible
解题思路：先打表计算四位数的素数，每次改变当前数的某一位，BFS，细节看代码
AC的C++代码：
#include<iostream>
#include<string>
#include<cstring>
#include<queue>
using namespace std;
const int N=10005;
bool prime[N+1];
bool vis[N];
void maketable(int n)
{
prime[0]=prime[1]=true;
for(int i=2;i<=n;i++)
if(!prime[i]){
for(int j=i+i;j<=n;j+=i)
prime[j]=true;
}
}
struct Node{
string v;
int step;
Node(){}
Node(string v,int step):v(v),step(step){}
};
int getval(string s)
{
int ans=0;
for(int i=0;i<s.length();i++)
ans=ans*10+s[i]-'0';
return ans;
}
int bfs(string s,string en)
{
memset(vis,false,sizeof(vis));
int v=getval(s);
vis[v]=true;
queue<Node>q;
q.push(Node(s,0));
while(!q.empty()){
Node f=q.front();
if(en==f.v)
return f.step;
q.pop();
for(int i=0;i<4;i++){
int st=(i==0)?1:0;
for(int j=st;j<=9;j++){
string str=f.v;
str[i]=j+'0';
int val=getval(str);
if(!vis[val]&&!prime[val]){//如果这个数是素数且没被访问
vis[val]=true;//标记这个数被访问了
q.push(Node(str,f.step+1));
}
}
}
}
return -1;//不存在就返回-1
}
int main()
{
int t;
string s,en;
cin>>t;
maketable(N);
while(t--){
cin>>s>>en;
int ans=bfs(s,en);
if(ans==-1)
cout<<"Impossible"<<endl;
else
cout<<ans<<endl;
}
return 0;
}