Prime Path
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 13813 | Accepted: 7796 |
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题意是说,给定两个四位素数a b 问从a变换到b,最少需要变换几次.变换的要求是,每次只能改变一个数字,而且中间过程得到的四位数也必须为素数.因为提到最少变换几次,容易想到bfs,bfs第一次搜到的一定是最短步数. 先打个素数表然后写个函数判断两个四位数有几位数字不同,如果只有一位,返回true,否则返回false然后竟然wa了两次!下表写错!pri[k++]=i;是先给pri[k]赋值,再k++;pri[++k]=i;才是先增加,再赋值.这个搞错了.所以wa了....sad
/*************************************************************************
> File Name: code/2015summer/searching/F.cpp
> Author: 111qqz
> Email: [email protected]
> Created Time: Fri 24 Jul 2015 01:16:23 AM CST
************************************************************************/
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)
typedef long long LL;
typedef unsigned long long ULL;
const int N =1E4+5;
int pri[N],which[N];
int a,b,k;
bool flag;
int d[N];
bool prime(int x)
{
for ( int i = 2 ; i*i<=x ;i++ )
{
if (x %i==0) return false;
}
return true;
}
bool ok (int x,int y)
{
if (d[y]!=-1) return false;
int res = 0; //记录两个数不对应不相等的数字的个数
int xx=x,yy=y;
while (x&&y)
{
if (x%10!=y%10) res++;
x = x/10;
y = y/10;
}
// if (res==1) cout<<"x:"<<xx<<" y:"<<yy<<endl;
if (res==1) return true;
return false;
}
void bfs()
{
queue<int>x;
memset(d,-1,sizeof(d));
x.push(a);
d[a]=0;
while (!x.empty())
{
int px = x.front();
// cout<<"px:"<<px<<endl;
x.pop();
if (px==b) return;
for ( int i = 0 ; i< k ; i++ )
{
if (ok(px,pri[i]))
{
d[pri[i]]=d[px]+1;
x.push(pri[i]);
}
}
}
}
int main()
{
k =0;
for ( int i = 1000; i <=9999; i++ )
{
if (prime(i))
{
pri[k++]=i;
}
}
int T;
// cout<<pri[0]<<endl;
// cout<<pri[1]<<endl;
cin>>T;
while (T--)
{
cin>>a>>b;
bfs();
if (d[b]==-1)
{
cout<<"Impossible"<<endl;
}
else
{
cout<<d[b]<<endl;
}
}
return 0;
}