搜索-I

• 原题

  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

• 思路解析
  题意：给定两个素数n和m，要求把n变成m，每次变换时只能变一个数字，即变换后的数与变换前的数只有一个数字不同，并且要保证变换后的四位数也是素数。求最小的变换次数；如果不能完成变换，输出Impossible。
  无论怎么变换，个位数字一定是奇数（个位数字为偶数肯定不是素数），这样枚举个位数字时只需枚举奇数就行；而且千位数字不能是0。所以可以用广搜，枚举各个数位上的数字，满足要求的数就加入队列，直到变换成功。因为是广搜，所以一定能保证次数最少。

• AC代码

#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
#define MAX 10000
#define DEBUG 1
bool isprime[MAX];
bool visit[MAX];
struct node
{
    int value;
    int step;
    node( int a, int b )
    {
        value=a;
        step=b;
    }
};
int pos[5]={1,10,100,1000,10000};
void sieve()
{
    memset(isprime,true,sizeof(isprime));
    isprime[0]=isprime[1]=false;
    for( int i=2;i<=100;i++ )
        for(int j=i;i*j<=MAX;j++)
            if(isprime[i])isprime[i*j]=false;
}
inline int digit(int a,int n)
{
    return (a%pos[n])/pos[n-1];
}
void bfs(int a,int b)
{
    queue<node>buffer;
    buffer.push(node(a,0));
    memset(visit,false,sizeof(visit));
    visit[a]=true;
    while(!buffer.empty())
    {
        node tmp=buffer.front();
        buffer.pop();
        if(tmp.value==b)
        {
            cout<<tmp.step<<endl;
            return;
        }
        for(int i=3;i>=0;i--)
        {
            node next=tmp;
            next.step++;
            int temp=next.value-digit(next.value,i+1)*pos[i];
            for( int j=0;j<=9;j++)
            {
                next.value=temp+j*pos[i];
                if(next.value==tmp.value||next.value<1000) continue;
                if(isprime[next.value] && !visit[next.value])
                {
                    buffer.push(next);
                    visit[next.value]=true;
                }
            }
        }
    }
    cout <<"Impossible"<<endl;
}
int main()
{
    sieve();
    int n,a,b;
    cin>>n;
    while(n--)
    {
        cin>>a>>b;
        bfs(a,b);
    }
    return 0;
}