[bfs]poj 3126 Prime Path

A - Prime Path

Time Limit:1000MS Memory Limit:65536KB

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

定义一个操作，是从一个四位素数到另一个四位素数的变换，其中变换条件为：素数内的数字只允许有一个发生变换。

给定input和output，求总共需要多少次操作才能从input变换到output。如果不能执行变换，输出Impossible。

很经典的dfs。定义一个s和step来存储数字和步骤数目，用vis[]来存储是否访问过该素数，用judge_prime()来判断是否为素数，把大部分不符合条件的数字给筛去。

先把input入队，step设置为0，然后将step+1，把其中符合的数字加入队列，依次扩展，直到发现符合的数字为止。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<queue>
#include<algorithm>
using namespace std;

int n, m;
const int N = 1e4 + 100;
int vis
;
struct node
{
int x, step;
};
queue<node> Q;

bool judge_prime(int x) //判断素数
{
if(x == 0 || x == 1)
return false;
else if(x == 2 || x == 3)
return true;
else
{
for(int i = 2; i <= (int)sqrt(x); i++)
if(x % i == 0)
return false;
return true;
}
}

void BFS()
{
int X, STEP, i;
while(!Q.empty())
{
node tmp;
tmp = Q.front();
Q.pop();
X = tmp.x;
STEP = tmp.step;
if(X == m)
{
printf("%d\n",STEP);
return ;
}
for(i = 1; i <= 9; i += 2) //个位
{
int s = X / 10 * 10 + i;
if(s != X && !vis[s] && judge_prime(s))
{
vis[s] = 1;
node temp;
temp.x = s;
temp.step = STEP + 1;
Q.push(temp);
}
}
for(i = 0; i <= 9; i++) //十位
{
int s = X / 100 * 100 + i * 10 + X % 10;
if(s != X && !vis[s] && judge_prime(s))
{
vis[s] = 1;
node temp;
temp.x = s;
temp.step = STEP + 1;
Q.push(temp);
}
}
for(i = 0; i <= 9; i++) //百位
{
int s = X / 1000 * 1000 + i * 100 + X % 100;
if(s != X && !vis[s] && judge_prime(s))
{
vis[s] = 1;
node temp;
temp.x = s;
temp.step = STEP + 1;
Q.push(temp);
}
}
for(i = 1; i <= 9; i++) //千位
{
int s = i * 1000 + X % 1000;
if(s != X && !vis[s] && judge_prime(s))
{
vis[s] = 1;
node temp;
temp.x = s;
temp.step = STEP + 1;
Q.push(temp);
}
}
}
printf("Impossible\n");
return ;
}

int main()
{
int t;
scanf("%d",&t);
while(t--)
{
while(!Q.empty()) Q.pop();
scanf("%d%d",&n,&m);
memset(vis,0,sizeof(vis));
vis
= 1;
node tmp;
tmp.x = n;
tmp.step = 0;
Q.push(tmp);
BFS();
}
return 0;
}