POJ 3216 Prime Path（打表+bfs）

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 27132		Accepted: 14861

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

Source

Northwestern Europe 2006

 

传送门（点我）

 

题目意思：

给出两个四位数的素数，要求对第一个素数进行一系列操作，使他变成第二个四位数的素数，每次操作只能改变一位数，且要求每次操作之后的数仍然是素数

问你最少的操作步骤数是多少？

 

分析：

先对四位数的素数打一个表

然后bfs

需要注意的是每次搜索入队之后，要进行回退操作

注意设置标记位（该数已经搜索过）

 

code：

#include<stdio.h>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <math.h>
#include <cstdlib>
#include <queue>
using namespace std;
#define max_v 10100
int prime[max_v];
int vis[max_v];
struct node
{
    int x,step;
};
void init()//打N以内的素数表，prime[i]=0为素数。
{
    memset(prime,,sizeof(prime));
    prime[]=;
    for(int i=; i<max_v; i++)
    {
        if(prime[i]==)
        {
            for(int j=; j*i<max_v; j++)
                prime[j*i]=;
        }
    }
}
int bfs(int s,int e)
{
    int num;
    memset(vis,,sizeof(vis));//vis[N]用来标记是否查找过
    queue<node> q;

    node p,next;

    p.x=s;
    p.step=;

    vis[s]=;
    q.push(p);

    while(!q.empty())
    {
        p=q.front();
        q.pop();

        if(p.x==e)
        {
            return p.step;
        }
        int t[];
        t[]=p.x/;//千
        t[]=p.x/%;//百
        t[]=p.x/%;//十
        t[]=p.x%;//个

        for(int i=; i<=; i++)
        {
            int temp=t[i];//这里特别注意，每次只能变换一位数，这里用来保存该变换位的数，变换下一位数时，该位数要恢复
            for(int j=; j<; j++)
            {
                if(t[i]!=j)
                {
                    t[i]=j;//换数
                    num=t[]*+t[]*+t[]*+t[];
                }
                if(num>=&&num<=&&vis[num]==&&prime[num]==)//满足要求，四位数，没有用过，是素数
                {
                    next.x=num;
                    next.step=p.step+;
                    q.push(next);
                    vis[num]=;
                }
            }
            t[i]=temp;//恢复
        }
    }
    return -;
}
int main()
{
    //题意：给你两个4位数，要求你每次只能变换一位数字，并且变换前后的数字都要满足是素数；求最少变换次数；
    //首先打好素数表，再进行BFS
    int n,a,b,ans;
    init();
    cin>>n;
    while(n--)
    {
        cin>>a>>b;
        ans=bfs(a,b);
        if(ans==-)
        {
            cout<<"Impossible"<<endl;
        }
        else
        {
            cout<<ans<<endl;
        }
    }
    return ;
}