POJ 3126 Prime Path （BFS）__搜尋

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.

1033173337333739377987798179

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

31033 81791373 80171033 1033

Sample Output

670

題意

給你n,m分別是素數，求由n到m變化的步驟數，規定每一步只能改變個十百千一位的數（千位不能為零），且變化得到的每一個數也為素數。

思路

首先進行素數打表，打出10000以內的所有素數（或者1000-10000），然後bfs。

對n變化個十百千的每一位，變化後檢測是否為素數，若是，排入佇列，同時標記已訪問，然後對隊列中的數進行同樣的操作，直到某一次變化產生m，輸出最少的變換次數。

若不可能達到m，則會在遍曆完[1000-9999]之間所有素數後退出。

AC 代碼

#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>using namespace std;#include<vector>#include<queue>#define Max 10000int prime[Max];bool isvis[Max];    //已經訪問過的數字不需要重新排入佇列void IsPrime()      //素數打表{    prime[0]=prime[1]=0;    prime[2]=1;    for(int i=3; i<Max; i++)        prime[i]=i%2==0?0:1;    int t=(int)sqrt(Max*1.0);    for(int i=3; i<=t; i++)        if(prime[i])            for(int j=i*i; j<Max; j+=2*i) //最佳化                prime[j]=0;}struct point{    int data;    int time;   //當前進行到第幾步    point(int data,int time)    {        this->data=data;        this->time=time;    }};int solve(point a,int b){    memset(isvis,false,sizeof(isvis));    queue<point>sk;    sk.push(a);    while(!sk.empty())    {        point p=sk.front();        sk.pop();        if(p.data==b)return p.time;        isvis[p.data]=true;        for(int i=10; i<=10000; i*=10)  //從低到高枚舉        {            for(int j=(i==10000)?1:0; j<=9; j++)    //最高位不能是0            {                int num=p.data/i*i+i/10*j+p.data%(i/10);    //計算當前的數字                if(!isvis[num]&&prime[num])                    sk.push(point(num,p.time+1));            }        }    }    return -1;}int main(){    IsPrime();    int n;    while(~scanf("%d",&n))    {        int a,b;        for(int i=0; i<n; i++)        {            scanf("%d%d",&a,&b);            int ans=solve(point(a,0),b);            printf(ans==-1?"Impossible\n":"%d\n",ans);        }    }    return 0;}