poj 3126 队列 且行且珍惜---

                                                Prime Path

                               Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

 

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

31033 81791373 80171033 1033

Sample Output

670

---

这道题是自己第二次打， 第一次的用的是count标记次数，那个是模仿大神写的所以有点不懂， 今天自己又用结构体打了一遍，虽然很艰难，但是收获确实很多的啊，这道题的基本题意就是让你由一个说变成下个数需要几步，每次只能改变数的一位，而且改变完的数必须是素数，所以首先要打素数表，然后一个一个列举出来所有的情况就行了，这道题的要求是找最短的方法，但是如果用队列来解这题的话就不用考虑了。因为用队列求出来的都是最短的。这道题我遇到的问题就是千位必须大于0要不然第二个测试数据的结果就变成5了所有我最后加了一个条件然后AC了。。。且行且珍惜---

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
int prime[10000];     //标记素数，非素数为1；
int v[10000];        //标记这个数在之前是否压入队列里
int ans,state;      //最后的步数和是否能找到那个数
int t[4];           //储存 千 百 十 个 
struct node
{
    int n;
    int step;
};
void Prime()           //打素数表
{
    int i, j;
    memset(prime,0,sizeof(prime));
    for(i=2;i<10000;i++)
    {
        if(!prime[i])
        {
            for(j=i+i;j<10000;j+=i)
                prime[j]=1;
        }
    }
}
void bfs(int a, int b)
{
    queue<node>q;
    struct node x;
    int i, j;
    x.n=a;
    x.step=0;
    q.push(x);
    if(x.n==b)
    {
        state=1;
        ans=x.step;
        return ;
    }
    memset(v,0,sizeof(v));
    while(!q.empty())
    {
        x=q.front();
        q.pop();
        if(x.n==b)
        {
            ans=x.step;
            break;
        }
        v[x.n]=1;
        t[0]=x.n/1000;
        t[1]=x.n/100%10;
        t[2]=x.n%100/10;
        t[3]=x.n%10;
        for(i=0;i<4;i++)
        {
            int m=t[i], sum;  ///一定把m存下来要不然就然后再j循环完后就达不到只改变一位的效果了
            for(j=0;j<10;j++)
            {
                t[i]=j;   
                sum=t[0]*1000+t[1]*100+t[2]*10+t[3];
                if(t[0]==0)
                    continue;
                if(!v[sum] && !prime[sum])
                {
                    struct node e;
                    e.n=sum;
                    e.step=x.step+1;
                    if(e.n==b)
                    {
                        state=1;
                        ans=e.step;
                        break;
                    }
                    q.push(e);
                    v[sum]=1;
                }
            }
            t[i]=m;
            if(state)
                break;
        }
        if(state)
            break;
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    Prime();
    prime[0]=1;prime[1]=1;
    while(t--)
    {
        int a, b;
        scanf("%d%d",&a,&b);
        ans=-1;state=0;
        bfs(a,b);
        if(ans==-1 && !state)
            printf("Impossible\n");
        else
            printf("%d\n",ans);
    }
}