poj3216 Prime Path（BFS）

Posted 2020-12-31 zhgyki

题目传送门

 Prime Path 

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
题意：给你四位数字a,b;每一不只能改变一位数字，且新的数字只能是素数，
要你输出最小步数

题解：bfs，每次向下遍历40个方向

代码：

#include<iostream>
#include<queue>
#include<string.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define mod 1000000007
#define INF 0x3f3f3f3f
struct niu
{
    int prime,step;
    niu(){}
    niu(int pr,int st)
    {
        prime=pr,step=st;
    }
};
int a,b;
int ans=INF;
bool check(int a)
{
    for(int i=2;i*i<=a;i++)
        if(a%i==0)return false;
    return a!=1;
}
queue<niu>q;
bool vis[10005];
int bfs()
{
    memset(vis,false,sizeof(vis));
    while(q.size())q.pop();
    q.push(niu(a,0));
    vis[a]=true;
    while(q.size())
    {
        niu tmp=q.front();q.pop();//cout<<tmp.prime<<tmp.step<<endl;
        if(tmp.prime==b)
        {
            ans=min(ans,tmp.step);
        }
        int cnt=tmp.prime%10;
        int cur=(tmp.prime/10)%10;
        for(int i=0;i<=9;i++)
        {
            int nx=(tmp.prime/10)*10+i;
            if(check(nx)&&!vis[nx])
            {
                vis[nx]=true;
                q.push(niu(nx,tmp.step+1));
            }
            int ny=(tmp.prime/100)*100+i*10+cnt;
            if(check(ny)&&!vis[ny])
            {
                vis[ny]=true;
                q.push(niu(ny,tmp.step+1));
            }
            int nz=(tmp.prime/1000)*1000+i*100+cur*10+cnt;
            if(check(nz)&&!vis[nz])
            {
                vis[nz]=true;
                q.push(niu(nz,tmp.step+1));
            }
            if(i==0)continue;
            int nn=(tmp.prime%1000)+i*1000;//cout<<nn<<endl;
            if(check(nn)&&!vis[nn])
            {
                vis[nn]=true;
                q.push(niu(nn,tmp.step+1));
            }
        }
    }
    return ans==INF?-1:ans;
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        ans=INF;//注意这里的ans要初始化
        cin>>a>>b;
        if(bfs()==-1)
            cout<<"Impossible"<<endl;
        else cout<<bfs()<<endl;
    }
    return 0;