POJ-3126-Prime Path【BFS】

3126-Prime Path

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

题目链接：POJ-3126

题目大意：

条件：四位数（没有前导0）且为素数

给出一个初始的数字a，每次能改变一位数字，使得这个新数字也满足条件，问最少需要几次到达b这个数字

题目思路：直接BFS就可以了，4位数字分别改变的情况（0 ~ 9）搜。需要注意中间数字也是要满足条件。

以下是代码：

//
//  F.cpp
//  搜索
//
//  Created by pro on 16/3/22.
//  Copyright (c) 2016年 pro. All rights reserved.
//

#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <string>
#include <cstring>
#include <queue>
using namespace std;
int vis[10000];
struct node
{
int val,cnt;
};
int prm[100000];   //记录各个素数
const int MAXV = 1e5;
bool isPrime[MAXV+1];   //判断该数字是不是素数
int size=0;   //表示prm数组的大小
void getPrime()
{
memset(isPrime, true, sizeof(isPrime));
int sq = sqrt((double)MAXV) + 1;
int i,j,k;
for(i = 2;i <= sq; i++)
if(isPrime[i])
for(j = 2,k = MAXV/i+1;j < k;j++)
isPrime[i*j] = false;
for( i = 2 ; i <= MAXV; i++)
if(isPrime[i])
prm[size++] = i;
isPrime[0] = isPrime[1] = false;
}
int getNum (int num,int i)
{
if (i == 1) return num % 10;
else if (i == 2) return num / 10 % 10;
else if (i == 3) return num / 100 % 10;
else return num / 1000;
}

int bfs(int begin,int end)
{
queue<node> que;
node zero;
zero.val = begin;
zero.cnt = 0;
que.push(zero);
vis[begin] = 1;
while(!que.empty())
{
node front = que.front();
que.pop();
if (front.val  == end) return front.cnt;
for (int i = 1; i <= 4; i++)
{
int num = getNum(front.val,i);
for (int j = 0; j < 10; j++)
{
if (j != num)
{
int new_num = front.val + (j - num) * pow(10,i - 1);
if (new_num < 1000) continue;
if (isPrime[new_num] && !vis[new_num])
{
vis[new_num] = 1;
node next;
next.val = new_num;
next.cnt = front.cnt + 1;
que.push(next);
}
}
}
}
}
return 0;
}

int main()
{
int t;
cin >> t;
getPrime();
while(t--)
{
memset(vis,0,sizeof(vis));
int begin,end;
cin >> begin >> end;
cout << bfs(begin,end) << endl;
}
}