hdoj--1016--Prime Ring Problem（递归回溯）

Problem Description      A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime.  Note: the number of first circle should always be 1.    Input      n (0 < n < 20). Output     The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order.     You are to write a program that completes above process.     Print a blank line after each case. Sample Input   6     8     Sample Output     Case 1:     1 4 3 2 5 6     1 6 5 2 3 4     Case 2:     1 2 3 8 5 6 7 4     1 2 5 8 3 4 7 6     1 4 7 6 5 8 3 2     1 6 7 4 3 8 5 2     本题就是考递归搜索的能力。     数据不大，其他Prime， map等的优化都没多大作用的。     记得记录好数据，就不会有问题了。     不过HDU的判断系统的确垃圾，其他OJ都不会在意末尾多一个空格或者回车的问题的，HDU就一个空格一个回车都一定要按照她的格式，否则就presentation error. #include const int MAX_N = 20; int num; int cycle[MAX_N]; bool vis[MAX_N]; bool isPrime(int n) {     for (int i = 2; i*i <= n; i++)     if (n % i == 0) return false;     return true; } bool isLegal(int val, int i) {     int left = i-1;     if (!isPrime(val+cycle[left])) return false;     if (i+1 == num && !isPrime(val+cycle[0])) return false;     return true; } void printNums(int i = 1) {      if (i == num)      {             for (int j = 0; j+1 < num; j++)          {             printf("%d ", cycle[j]);         }         printf("%d\n", cycle[i-1]);         return ;     }     for (int v = 2; v <= num; v++)     {         if (vis[v]) continue;         if (isLegal(v, i))         {             cycle[i] = v;             vis[v] = true;             printNums(i+1);             vis[v] = false;         }     } } int main() {     int t = 0;     cycle[0] = 1;     while (scanf("%d", &num) != EOF)     {         printf("Case %d:\n", ++t);         for (int i = 2; i <= num; i++) vis[i] = false;         printNums();         putchar('\n');     }     return 0;     }