Prime Path POJ 3126

题目链接：http://poj.org/problem?id=3126

题意：现给出一个四位数，要求每次只能改变个十百千中的一位，在每次改变之后这个这个数必须是素数，求改变成指定数字的最少改变次数。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <cstring>
using namespace std;
#define INF 0xfffffff
#define maxn 10000
int p[maxn], v[maxn];

struct node
{
    int x, step;
};

int prime(int n)
{
    int i, k=sqrt(n);

    for(i=2; i<=k; i++)
        if(n %i == 0)
        return 0;

    return 1;
}


/*
  学长的Turn:
  
  int Turn(int n, int k)
{
    char s[10]={0};

    sprintf(s, "%d", n);
    s[k] = '0';
    sscanf(s, "%d", &n);

    return n;

}
*/

int Turn(int n, int k)   
{
    int a, b, c;

     if(k==0)
     {
         n %= 1000;
     }
     else if(k==1)
     {
          a = n /1000;
          b = n %1000;
          c = b %100;
          n = a*1000+c;
     }
     else if(k==2)
     {
         a = n/1000;
         b = n%1000;
         c = b/100;
         b = b%10;
         n = a*1000+c*100+b;
     }
     else
     {
         a = n / 10;
         n = a * 10;
     }

    return n;
}

int BFS(int x, int y)
{
    queue<node>Q;
    node begins, now, next;
    begins.x = x;
    begins.step = 0;

    Q.push(begins);

    while(Q.size())
    {
        now = Q.front();
        Q.pop();

        if(now.x == y)  return now.step;

        int t = 1000, q;

        for(int i=0; i<4; i++)
        {
            q = Turn(now.x, i);

            for(int j=0; j<10; j++)
            {
                next.x = q + j * t;

                if(p[next.x] && !v[next.x])
                {
                    next.step = now.step + 1;
                    v[next.x] = 1;
                    Q.push(next);
                }
            }

            t /= 10;
        }
    }
    return -1;
}

int main()
{
    int i, T, s, e;

    memset(p, 0, sizeof(p));

    for(i=1000; i<maxn; i++)
        p[i] = prime(i);

    scanf("%d", &T);

    while(T --)
    {
        scanf("%d %d", &s, &e);

         memset(v, 0, sizeof(v));

        int ans = BFS(s, e);

        if(ans == -1)  printf("Impossible
");
        else  printf("%d
", ans);
    }
    return 0;
}