F - 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

Output

Sample Input
3
1033 8179
1373 8017
1033 1033

Sample Output
6
7
0

程式碼實現

#include <stdio.h>
#include <math.h>
#include <queue>
#include <string.h>
using namespace std;

const int MAX=10050;
int su[MAX],book[MAX],num[4],m,n,temp,vis[MAX];  //book記錄數字是否已經入隊  //nun陣列記錄各個位的值 例如1234 記錄1 2 3 3

//
struct  node{        //定義結構體儲存當前的數字和路徑
    int x,step;
}en,st;
queue<node>q;
int su_t( int n){    //判斷素數 若為素數的話返回1  否則返回0
    int i,t;
    t=sqrt(n+0.0);   //開根號，(n+0.0)轉換為double型別
    for (i=2;i<=t;i++){
        if (n%i==0) return 0;
    }
    return 1;
}


int main() {
    for(int i=2;i<MAX;i++) su[i]=su_t(i);    //在組數中標記下標是否為素數，素數存1 非素數存0
    int t;
    scanf("%d",&t);
    while(t--){
        memset(book,0, sizeof(book));    //初始化是否已經入隊
        scanf("%d %d",&m,&n);
        if(m==n) {
            printf("0\n");               //特判，如果兩數相等，則返回0
            continue;
        }

        memset(vis,-1, sizeof(vis));   //初始化路徑記錄值

        flag=0;
        while(!q.empty()) q.pop();    //佇列清空
        st.x=m;
        st.step=0;
        book[st.x]=1;
        vis[st.x]=0;
        q.push(st);                    //頭入佇列
        while(!q.empty()){
            st=q.front();
            q.pop();
            book[st.x]=0;            //標記隊頭已經入隊
            temp=st.x;
            for(int i=0;i<4;i++){   //記錄該數各個位上的值
                num[i]=temp%10;
                temp/=10;
            }
            en.step=st.step+1;
            for(int i=0;i<=9;i++){        //換掉個位上的數
                en.x=i+num[1]*10+num[2]*100+num[3]*1000;
                if(vis[en.x]==-1 || vis[en.x]>en.step){  //如果該處vis值為-1或該處儲存的路徑值較大，則更新它
                    vis[en.x]=en.step;
                    if(su[en.x]==1&& !book[en.x]){  //如果滿足條件則入隊
                        book[en.x]=1;  //把入隊的數標記為1
                        q.push(en);   //入隊
                    }
                }

            }

            for(int i=0;i<=9;i++){      //更換十位值 ,詳解參考上面註釋
                en.x=num[0]+i*10+num[2]*100+num[3]*1000;
                if(vis[en.x]==-1 || vis[en.x]>en.step){
                    vis[en.x]=en.step;
                    if(su[en.x]==1 && !book[en.x]){
                        book[en.x]=1;
                        q.push(en);
                    }
                }

            }

            for(int i=0;i<=9;i++){   //更換百位值,詳解參考上面註釋
                en.x=num[0]+num[1]*10+i*100+num[3]*1000;
                if(vis[en.x]==-1 || vis[en.x]>en.step){
                    vis[en.x]=en.step;
                    if(su[en.x]==1 && !book[en.x]){
                        book[en.x]=1;
                        q.push(en);
                    }

                }

            }

            for(int i=1;i<=9;i++){    //更換千位值,詳解參考上面註釋
                en.x=num[0]+num[1]*10+num[2]*100+i*1000;
                if(vis[en.x]==-1 || vis[en.x]>en.step){
                    vis[en.x]=en.step;
                    if(su[en.x]==1 && !book[en.x]) {
                        book[en.x]=1;
                        q.push(en);
                    }
                }


            }

        }

        printf("%d\n",vis[n]);  

    }
    return 0;
}