广搜 poj 3126

Prime Path

Time Limit: 1000MS Memory Limit: 65536K

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

题目大意就是输入两个四位素数数，每次改变前面的素数的一位，改变后的数也要是素数，然后再改变变后的苏素数的一位。问最小多少步可以变成后面那个素数。例如样例1：

   1033-->1733-->3733-->3739-->3779-->8779-->8179  一共六步。。

 这一题用广搜。同时用素数表来判断一个数是否为素数。。模拟操作的过程（改变个位、改变百位……）。

代码：

#include <iostream>
#include <queue>
#include <string.h>
using namespace std;

bool visit[10000];   //判断i（visit[i]）是否已经被访问过。
bool prime[10000];   //判断i (prime[i]) 是否为素数。
int cost[10000];     //判断变成 i (cost[i])时 花费多少钱。

int bfs(int x,int y)
{
   queue<int> q;     //建立一个int型的队列。
   q.push(x);         //将x压入队列。
   visit[x]=true;       //初始化 x被访问过。
   while(!q.empty())
   {
	   int now=q.front();       //将q的队头赋值给now。
	   q.pop();
	   for(int i=0;i<=9;i++)
	   {
		   int temp=(now/10)*10+i;        //改变个位。
		   if(prime[temp]==true && visit[temp]==false)
		   {
               visit[temp]=true;
			   q.push(temp);
			   cost[temp]=cost[now]+1;
		   }

		   temp=(now/100)*100+i*10+now%10;  //改变十位。
		   if(prime[temp]==true && visit[temp]==false)
		   {
			   visit[temp]=true;
			   q.push(temp);
			   cost[temp]=cost[now]+1;
		   }

		   temp=(now/1000)*1000+i*100+now%100;          //改变千位。
		   if(prime[temp]==true && visit[temp]==false)
		   {
			   visit[temp]=true;
			   q.push(temp);
			   cost[temp]=cost[now]+1;
		   }

		   temp=now%1000+i*1000;            //改变万位。
		   if(i!=0 && prime[temp]==true && visit[temp]==false)
		   {
			   visit[temp]=true;
			   q.push(temp);
			   cost[temp]=cost[now]+1;
		   }

		   if(visit[y])     //表示变成了y.
		      return cost[y]; 
	   }

   }
   return 0;
}

int main()
{
	int i,j,k;
	memset(prime,true,sizeof(prime));
	for( i=2;i<=10000;i++)           //建立素数表。
	{
		if(prime[i])
		{
		  for(int j=i;j<=10000/i;j++)
		 	prime[i*j]=false;      //这个很巧妙，排除不是素数的，true表示是素数。
		}
	}
	int n,num1,num2;
	cin>>n;
	for(i=0;i<n;i++)
	{
		memset(visit,false,sizeof(visit));
		memset(cost,0,sizeof(cost));
       cin>>num1>>num2;
	   cout<<bfs(num1,num2)<<endl;
	}
	return 0;
}