POJ3126Prime Path 暴力BFS

POJ3126Prime Path 暴力BFS

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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：给定两个数，要求从a变到b每次只能变一位数，而且变完之后的数必须是素数（这是控制越界的条件），一开始我并没什么想法,直到zmin提醒到我，暴力，暴力，暴力，暴力枚举每一位，提醒了我，但是还是有一点需要注意的就是，注意在变每一位的时候把他存下来，方便枚举下一位的时候恢复,提醒自己 素数表千万不要打错，素数表千万不要打错，素数表千万不要打错。

////  main.cpp//  whr_bfs////  Created by  41 on 17/8/7.//  Copyright (c) 2017年 henuwhr. All rights reserved.//  赵晓宇说我要ac#include <iostream>#include <cstdio>#include <algorithm>#include <queue>#include <map>#include <cstring>using namespace std;const int maxx = 20005;int vis[maxx];int prime[maxx];int a,b;void init() ///打N以内的素数表，prime[i]=0为素数。{    int i,j;    memset(prime,0,sizeof(prime));    prime[1]=1;    for(i=2; i<maxx; i++)    {        if(prime[i]==0)        {            for(j=2; j*i<maxx; j++)                prime[j*i]=1;        }    }}struct node{    int x,st;};void bfs(int x){    init();    queue<node>q;    node xx,nex;    xx.x = x;    xx.st = 0;    q.push(xx);    vis[xx.x] = 1;    while(!q.empty()){        node p;        p = q.front();        q.pop();        if(p.x==b){            cout<<p.st<<endl;            return ;        }        int t[6];        t[1]=p.x/1000;        t[2]=p.x/100%10;        t[3]=p.x/10%10;        t[4]=p.x%10;        int mm=0;        for(int i = 1 ; i<=4;i++){            int xxxx = t[i];//保存原来的。以后方便恢复            for(int j = 0;j<=9;j++){                if(t[i]!=j){                    t[i] = j;                    mm = t[1]*1000+t[2]*100+t[3]*10+t[4];                }                if(mm>=1000&&mm<10000&&!vis[mm]&&!prime[mm]){                    nex.x = mm;                    nex.st = p.st+1;                    q.push(nex);                    vis[mm] = 1;                }            }            t[i] = xxxx;        }    }    //cout << "Impossible" << endl;}int main(int argc, const char * argv[]) {    ios::sync_with_stdio(false);    int t;    cin>>t;    while(t--){        cin>>a>>b;        memset(vis,0,sizeof vis);        bfs(a);    }    return 0;}/* 3 1033 8179 1373 8017 1033 1033 *//* 6 7 0 */