POJ 3126 Prime Path_BFS

注意结构体队列的使用，在队列中，一条完整的路径才是正确的，中间的歧路最后还是走不通

转载：http://blog.csdn.net/yulanarti/article/details/1787971  ——> 说明sprintf()的作用

注意 sprintf(num, "%d", cur.num);//将数字转化为字符串，可以替代 itoa

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cstdlib>
  5 #include <cmath>
  6 #include <queue>
  7 #define sc(x,y) scanf("%d %d",&(x),&(y))
  8 #define pf(x) printf("%d\n", x)
  9 #define CL(x, y) memset(x,y,sizeof(x))
 10 using namespace std;
 11 const int MAX = 100000;
 12 int prime[MAX] = {0};   //素数表
 13 int N, a, b, i, j, temp;
 14 int used[MAX];
 15 void InitPrime();
 16 void BFS(int x, int y);
 17 struct node
 18 {
 19     int num;
 20     int step;
 21 };
 22 queue <node> Q;
 23 int main()
 24 {
 25     InitPrime();
 26     cin >> N;
 27     while(N--)
 28     {
 29         cin >> a >> b;
 30         CL(used, 0);
 31         BFS(a, b);
 32     }
 33     return 0;
 34 }
 35 void BFS(int front, int rear)
 36 {
 37     while(!Q.empty())
 38         Q.pop();//Q.clear();清空
 39     node first, cur, next;
 40     first.num = front;
 41     first.step = 0;
 42     used[front] = 1;
 43     Q.push(first);
 44     while(!Q.empty())
 45     {
 46         cur = Q.front();
 47         Q.pop();
 48         if(cur.num == rear)
 49         {
 50             cout << cur.step << endl;
 51             return ;
 52         }
 53         for(i = 0; i < 4; i++)
 54         {
 55             char num[5];
 56             sprintf(num, "%d", cur.num);//将数字转化为字符串，可以替代 itoa
 57             for (j = 0; j < 10; j++)
 58             {
 59                 if (j == 0 && i == 0)//千位不能为 0
 60                     continue;
 61                 if(i == 0)
 62                     temp = j*1000+(num[1]-‘0‘)*100+(num[2]-‘0‘)*10+(num[3]-‘0‘);//千位
 63                 else if(i == 1)
 64                     temp = j*100+(num[0]-‘0‘)*1000+(num[2]-‘0‘)*10+(num[3]-‘0‘);//百位
 65                 else if(i == 2)
 66                     temp = j*10+(num[0]-‘0‘)*1000+(num[1]-‘0‘)*100+(num[3]-‘0‘);//十位
 67                 else if(i == 3)
 68                     temp = j+(num[0]-‘0‘)*1000+(num[1]-‘0‘)*100+(num[2]-‘0‘)*10;//个位
 69                 if(prime[temp] && !used[temp])
 70                 {
 71                     next.num = temp;
 72                     next.step = cur.step+1;
 73                     used[temp] = 1;
 74 //                    cout << next.num << " " << next.step << endl;
 75                     Q.push(next);
 76                     Q.push(next);
 77                 }
 78             }
 79         }
 80     }
 81     printf("Impossible\n");
 82     return ;
 83 }
 84 void InitPrime()
 85 {
 86     int i, j;
 87     for(i = 1009; i <= 9973; i++)
 88     {
 89         if(!(i&1))continue;  //奇偶判断
 90         for(j = 3; j*j <= i; j += 2)
 91         {
 92             if(0 == i%j)
 93             {
 94                 break;
 95             }
 96         }
 97         if(j * j > i)
 98         {
 99             prime[i] = 1;
100         }
101     }
102 }