Sicily 1444: Prime Path（BFS）

　　题意为给出两个四位素数A、B，每次只能对A的某一位数字进行修改，使它成为另一个四位的素数，问最少经过多少操作，能使A变到B。可以直接进行BFS搜索

 #include<bits/stdc++.h>
 using namespace std;

 bool isPrime(int n){//素数判断
     if(n ==  || n == ) return true;
     else{
         int k = sqrt(n) + ;
         for(int i = ; i < k; i++){
             if(n % i == ) return false;
         }
         return true;
     }
 }

 bool Prime[];
 int visit[];
 void getPrime(){
     for(int i = ; i < ; i++){
         if(isPrime(i))Prime[i] = true;
     }
 }

 struct Node{
     int num;//储存数字
     int cost;//操作步数
 }; 

 Node bfs(int start, int end){
     queue<Node> q;
     Node front;
     front.num = start;
     front.cost = ;
     visit[start] = ;
     q.push(front);
     while(!q.empty()){
         front = q.front(); q.pop();

         for(int i = ; i < ; i++){//换千位
             int m = front.num;
             m = m %  + i * ;
             if(!visit[m] && Prime[m]){
                 visit[m] = ;
                 Node tmp = front;
                 tmp.num = m;
                 tmp.cost++;
                 q.push(tmp);

                 if(m == end)return tmp;
             }
         }

         for(int i = ; i < ; i++){//换百位
             int m = front.num;
             m = m %  + (m/) *  + i * ;
             if(!visit[m] && Prime[m]){
                 visit[m] = ;
                 Node tmp = front;
                 tmp.num = m;
                 tmp.cost++;
                 q.push(tmp);

                 if(m == end)return tmp;
             }
         }

         for(int i = ; i < ; i++){//换十位
             int m = front.num;
             m = m %  + (m/) *  + i * ;
             if(!visit[m] && Prime[m]){
                 visit[m] = ;
                 Node tmp = front;
                 tmp.num = m;
                 tmp.cost++;
                 q.push(tmp);
                 if(m == end)return tmp;
             }
         }

         for(int i = ; i < ; i++){//换个位
             int m = front.num;
             m = (m/) *  + i;
             if(!visit[m] && Prime[m]){
                 visit[m] = ;
                 Node tmp = front;
                 tmp.num = m;
                 tmp.cost++;
                 q.push(tmp);

                 if(m == end)return tmp;
             }
         }
     }
     Node tmp;
     tmp.num = ;
     tmp.cost = ;
     return tmp;
 }
 int main(){
     int n;
     cin >> n;
     getPrime();
     while(n--){
         int a, b;
         cin >> a >> b;
         Node tmp;
         memset(visit, , sizeof(visit));
         tmp = bfs(a, b);
         if(tmp.num ==  && tmp.cost == ) cout <<  << endl;
         else{
             cout << tmp.cost << endl;
         }
     }
 }