[kuangbin带你飞]专题一 简单搜索 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
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

用两种处理这个四位数，至今不知道为何之前wa两次，后来想起来用sprintf可以方便的提取各个位数。然后就A了，结论就是尽量简洁的代码，出错的可能性就越少。

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
bool visited[10010];
bool prime[10010]={false,false,false};
int goal;

int init(){
    for(int i = 2;i< 10000;i++){
        if(!prime[i]){
            for(int j = 2 ; i*j<10000;j++){
                prime[j*i] = 1; 
             } 
        }
    }
}

struct node{
    int num;
    int step;
};

int bfs(node start){
    queue<node> Q;
    Q.push(start);
    node nownode;
    node nextnode;
    while(!Q.empty()){
        char num[5];
        nownode = Q.front();Q.pop();
        if(nownode.num == goal){
            return nownode.step;
            break;
        }
        sprintf(num,"%d",nownode.num);
        for(int i=0;i<4;i++){
            for(int j=0;j<=9;j++){
                if(i==0&&j==0) continue; 
                switch(i){
                    case 0:
                            nextnode.num = j*1000 + (num[1]-'0')*100+(num[2]-'0')*10 +(num[3]-'0'); break;
                    case 1: 
                            nextnode.num = (num[0]-'0')*1000 + j*100+(num[2]-'0')*10 +(num[3]-'0'); break;
                    case 2:
                            nextnode.num = (num[0]-'0')*1000 + (num[1]-'0')*100+j*10 +(num[3]-'0'); break;
                    case 3:
                            nextnode.num = (num[0]-'0')*1000 + (num[1]-'0')*100+(num[2]-'0')*10 +j; break;      
                }
                
                if(!visited[nextnode.num]&&!prime[nextnode.num]){
//                  cout<<"next"<<nextnode.num<<endl;
//              cout<<"now"<<nownode.num<<endl;
                    visited[nextnode.num]=true;
                    nextnode.step = nownode.step + 1;
                    
                    Q.push(nextnode); 
                }
            } 
        }
        nextnode.step = nownode.step ++;
        
    }
    
    return -1;
    
}
int main() {
    int T;
    cin>>T;
    init();
    while(T--){
        node start;
        cin>>start.num>>goal;
        memset(visited,false,sizeof(visited));
        start.step = 0;
        int ans = bfs(start);
        if(ans == -1)
            cout<<"Impossible"<<endl;
        else
            cout<<ans<<endl;    
    }   
    return 0;
}