POJ 3126-Prime Path（BFS-素数变换一位）

最新推荐文章于 2019-08-24 11:27:19 发布

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最新推荐文章于 2019-08-24 11:27:19 发布

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分类专栏： POJ 初级计划 ACM_搜索文章标签： POJ 3126 Prime Path BFS 素数变换一位素数

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本文介绍了一个有趣的算法问题——素数路径。任务要求从一个四位素数出发，通过改变一位数字到达另一个四位素数，且每一步都保持素数性质。文章详细解释了使用BFS进行搜索的过程，并给出了具体实现代码。

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Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19095		Accepted: 10711

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

Sample Output

6
7
0

Source

Northwestern Europe 2006

题目意思：

给两个是素数的四位数（千位始终不为0），每次改变其中一位，使得改变后的数依然是素数，计算从前一个数到后一个数最少一共需要改变几次。

解题思路：

打表1000~9999中所有的素数。

BFS：队列先将第一个数入队，然后分解出其各个位上的数，循环依次枚举个位~千位，可以是0~9中的任意一个数。

注意每次位数循环维护当前位不变，保证下次循环仅改变下一位上的数。

剪枝：①枚举到与当前位上的数相同的数，②处理过的数。

#include<iostream>
#include<cstdio>
#include<iomanip>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<map>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define INF 0x3f3f3f3f
#define MAXN 10000
short int d[4];
bool prime[MAXN],vis[MAXN];
int cnt[MAXN];
void init()//1000~9999中所有的素数
{
    memset(prime,false,sizeof(prime));
    int i,j;
    for(i=1000; i<=10000; i++)
    {
        for(j=2; j<i; j++)
            if(i%j==0)
            {
                prime[i]=false;
                break;
            }
        if(j==i) prime[i]=true;
    }
}
int bfs(int a,int b)
{
    queue <int> q;
    q.push(a);
    vis[a]=true;
    while(!q.empty())
    {
        int s=q.front();
        if(s==b) return cnt[s];
        q.pop();
        d[0]=s%10;//个位
        d[1]=s%100/10;//十位
        d[2]=s%1000/100;//百位
        d[3]=s/1000;//千位
        /*for(int j=3; j>=0; --j)
            cout<<d[j]<<" ";
        cout<<endl;*/
        int temp,x;
        for(int j=0; j<4; ++j) //个位~千位
        {
            x=d[j];//记录当前位置改变之前的数
            for(int i=0; i<10; ++i) //0~9
            {
                if(x==i) continue;//与当前位相同，剪枝
                if(j==3&&i==0)continue;//千位不能为0
                d[j]=i;//改变当前位的数
                temp=d[3]*1000+d[2]*100+d[1]*10+d[0];
                if(vis[temp]) continue;//已访问，剪枝
                if(prime[temp])//是素数
                {
                    q.push(temp);
                    vis[temp]=true;
                    cnt[temp]=cnt[s]+1;//记录变化次数
                }
                if(temp==b) return cnt[temp];
            }
            d[j]=x;//维护当前位不变，保证下次循环仅改变下一位
        }
    }
    return -1;
}
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("F:/cb/read.txt","r",stdin);
    //freopen("F:/cb/out.txt","w",stdout);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin>>t;
    init();
    while(t--)
    {
        memset(vis,false,sizeof(vis));
        memset(cnt,0,sizeof(cnt));
        int a,b,ans=0;
        cin>>a>>b;
        ans=bfs(a,b);
        if(ans==-1) cout<<"Impossible"<<endl;
        else cout<<ans<<endl;
    }
    return 0;
}
/*
3
1033 8179
1373 8017
1033 1033
*/