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CD0J/POJ 851/3126 方老师与素数/Prime Path BFS

Prime Path
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 9982 Accepted: 5724

Description

技术分享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

31033 81791373 80171033 1033

Sample Output

670


#include <cstdio>#include <cmath>#include <cstring>#include <ctime>#include <iostream>#include <algorithm>#include <set>#include <vector>#include <sstream>#include <queue>#include <typeinfo>#include <fstream>typedef long long ll;using namespace std;//freopen("D.in","r",stdin);//freopen("D.out","w",stdout);#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)#define maxn 100001const int inf=0x7fffffff;   //无限大const int MAXN = 10000;bool flag[MAXN];int primes[MAXN], pi;struct point{    int x;    int y;};void GetPrime_1(){    int i, j;    pi = 0;    memset(flag, false, sizeof(flag));    for (i = 2; i < MAXN; i++)        if (!flag[i])        {            primes[i] = 1;//素数标识为1            for (j = i; j < MAXN; j += i)                flag[j] = true;        }}int vis[maxn];int main(){    GetPrime_1();    int t;    cin>>t;    while(t--)    {        memset(vis,0,sizeof(vis));        int n,m;        cin>>n>>m;        vis[n]=1;        queue<point> q;        q.push((point){n,0});        int flag1=0;        while(!q.empty())        {            point now=q.front();            if(now.x==m)            {                flag1=now.y;                break;            }            point next;            for(int i=0;i<=9;i++)            {                next.x=now.x/10;                next.x*=10;                next.x+=i;                next.y=now.y+1;                if(next.x<1000||next.x>=10000)                    continue;                if(vis[next.x]==1)                    continue;                if(next.x==m)                {                    flag1=next.y;                    break;                }                if(primes[next.x]==1)                {                    //cout<<next.x<<endl;                    vis[next.x]=1;                    q.push(next);                }            }            for(int i=0;i<=9;i++)            {                int temp=now.x%10;                next.x=now.x/100;                next.x*=100;                next.x+=i*10;                next.x+=temp;                if(next.x<1000||next.x>=10000)                    continue;                if(vis[next.x]==1)                    continue;                if(next.x==m)                {                    flag1=next.y;                    break;                }                if(primes[next.x]==1)                {                    //cout<<next.x<<endl;                    vis[next.x]=1;                    q.push((point){next.x,now.y+1});                }            }            for(int i=0;i<=9;i++)            {                int temp=now.x%100;                next.x=now.x/1000;                next.x*=1000;                next.x+=i*100;                next.x+=temp;                if(next.x<1000||next.x>=10000)                    continue;                if(vis[next.x]==1)                    continue;                if(next.x==m)                {                    flag1=next.y;                    break;                }                if(primes[next.x]==1)                {                    //cout<<next.x<<endl;                    vis[next.x]=1;                    q.push((point){next.x,now.y+1});                }            }            for(int i=0;i<=9;i++)            {                int temp=now.x%1000;                next.x=now.x/10000;                next.x*=10000;                next.x+=i*1000;                next.x+=temp;                if(next.x<1000||next.x>=10000)                    continue;                if(vis[next.x]==1)                    continue;                if(next.x==m)                {                    flag1=next.y;                    break;                }                if(primes[next.x]==1)                {                //    cout<<next.x<<endl;                    vis[next.x]=1;                    q.push((point){next.x,now.y+1});                }            }            if(flag1>0)            break;            q.pop();        }    printf("%d\n",flag1);    }    return 0;}

 

CD0J/POJ 851/3126 方老师与素数/Prime Path BFS