Sicily 1444: Prime Path（BFS）

Posted 2020-08-26 Vincent_Bryan

　　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

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 
  4 bool isPrime(int n){//素数判断 
  5     if(n == 2 || n == 3) return true;
  6     else{
  7         int k = sqrt(n) + 1;
  8         for(int i = 2; i < k; i++){
  9             if(n % i == 0) return false;
 10         }
 11         return true;
 12     }
 13 }
 14 
 15 bool Prime[10010]; 
 16 int visit[10010];
 17 void getPrime(){
 18     for(int i = 1000; i < 10000; i++){
 19         if(isPrime(i))Prime[i] = true;
 20     }
 21 }
 22 
 23 struct Node{
 24     int num;//储存数字 
 25     int cost;//操作步数 
 26 }; 
 27 
 28 Node bfs(int start, int end){
 29     queue<Node> q;
 30     Node front;
 31     front.num = start;
 32     front.cost = 0;
 33     visit[start] = 1;
 34     q.push(front);
 35     while(!q.empty()){
 36         front = q.front(); q.pop();
 37         
 38         for(int i = 1; i < 10; i++){//换千位 
 39             int m = front.num;
 40             m = m % 1000 + i * 1000;
 41             if(!visit[m] && Prime[m]){
 42                 visit[m] = 1; 
 43                 Node tmp = front;
 44                 tmp.num = m;
 45                 tmp.cost++;
 46                 q.push(tmp);
 47                 
 48                 if(m == end)return tmp; 
 49             }
 50         }
 51         
 52         for(int i = 0; i < 10; i++){//换百位 
 53             int m = front.num;
 54             m = m % 100 + (m/1000) * 1000 + i * 100;
 55             if(!visit[m] && Prime[m]){
 56                 visit[m] = 1; 
 57                 Node tmp = front;
 58                 tmp.num = m;
 59                 tmp.cost++;
 60                 q.push(tmp);
 61                 
 62                 if(m == end)return tmp; 
 63             }
 64         }
 65         
 66         for(int i = 0; i < 10; i++){//换十位 
 67             int m = front.num;
 68             m = m % 10 + (m/100) * 100 + i * 10;
 69             if(!visit[m] && Prime[m]){
 70                 visit[m] = 1; 
 71                 Node tmp = front;
 72                 tmp.num = m;
 73                 tmp.cost++;
 74                 q.push(tmp);
 75                 if(m == end)return tmp; 
 76             }
 77         }
 78         
 79         for(int i = 0; i < 10; i++){//换个位 
 80             int m = front.num;
 81             m = (m/10) * 10 + i;
 82             if(!visit[m] && Prime[m]){
 83                 visit[m] = 1; 
 84                 Node tmp = front;
 85                 tmp.num = m;
 86                 tmp.cost++;
 87                 q.push(tmp);
 88                 
 89                 if(m == end)return tmp; 
 90             }
 91         }
 92     }
 93     Node tmp;
 94     tmp.num = 0;
 95     tmp.cost = 0;
 96     return tmp;
 97 }
 98 int main(){
 99     int n;
100     cin >> n;
101     getPrime();
102     while(n--){
103         int a, b;
104         cin >> a >> b;
105         Node tmp;
106         memset(visit, 0, sizeof(visit));
107         tmp = bfs(a, b);
108         if(tmp.num == 0 && tmp.cost == 0) cout << 0 << endl;
109         else{
110             cout << tmp.cost << endl;
111         }
112     } 
113 }