POJ3126——BFS——Prime Path

http://poj.org/problem?id=3126

大意：从一个素数到另一个素数，每次只能改变一个数，要求每次改变后的数都是素数，要求算出转换的最小次数，简单BFS，不过注意d要还原

/************************************************
* Author        :Powatr
* Created Time  :2015-8-9 16:09:31
* File Name     :F.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAXN = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[5], b[5];
int vis[110150];
queue<pair<int, int> > q;
int prim[21015];
void inti()
{
    for(int i = 1000; i < 10000; i++){
        for(int j = 2; j <i/2; j++){
            if(i%j == 0)  {prim[i] = 1;break;}
            }
        }

}
int main(){
    int T;
    int n, m;
    memset(prim, 0, sizeof(prim));
    scanf("%d", &T);
    inti();
    while(T--){
        memset(vis, 0, sizeof(vis));
        scanf("%d%d", &n, &m);
        a[1] = n/1000, a[2] = n%1000/100, a[3] = n%100/10, a[4] = n%10;
        while(!q.empty()) q.pop();
        q.push(make_pair(n, 0));
        while(!q.empty()){
        int x = q.front().first, y = q.front().second;
        q.pop();
        if(vis[x]) continue;
        vis[x] = 1;
       // printf("%d
", x);
        a[1] = x/1000, a[2] = x%1000/100, a[3] = x%100/10, a[4] = x%10;
        if(x == m){
              printf("%d
", y);
              break;
        }
                for(int i = 1; i <= 9; i++){
                int k = i*1000 + a[2]*100 + a[3]*10 + a[4];
                if(!prim[k])
                q.push(make_pair(k,y+1));
                }
                for(int i = 0; i <= 9; i++){
               int k = a[1]*1000 + i*100 + a[3]*10 + a[4];
               if(!prim[k])
               q.push(make_pair(k, y+1));
                } 
                for(int i = 0 ; i <= 9; i++){
                int k = a[1]*1000 + a[2]*100 + i*10 + a[4];
                if(!prim[k])
                q.push(make_pair(k,y+1));
                }
                for(int i = 0; i <= 9; i++){
                int k = a[1]*1000 + a[2]*100 + a[3]*10 + i;
                if(!prim[k])
                q.push(make_pair(k,y+1));
                }
        }
    }
        return 0;
}