Prime Ring Problem HDU - 1016 (dfs)
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Prime Ring Problem
HDU - 1016
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.
Note: the number of first circle should always be 1.
Inputn (0 < n < 20).
OutputThe 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
1 #include<iostream> 2 #include<stdio.h> 3 #include<cstring> 4 #include<queue> 5 6 using namespace std; 7 8 9 int n; 10 int a[23]; 11 int vis[23]; 12 bool mark[1000]; 13 14 void init() // 素数筛 15 16 for(int i = 2; i < 1000; ++i) 17 mark[i] = true; 18 19 for(int i = 2; i < 1000; ++i) 20 21 if(mark[i] == true) 22 23 for(int j = i*i; j < 1000; j += i) 24 25 mark[j] = false; 26 27 28 29 30 31 // 边枚举边判断,不要最后一次性判断,会超时 32 void dfs(int step) 33 34 if(step > 2) 35 36 if(mark[a[step-1]+a[step-2]] == false) // 判断最后两个数 37 return; 38 39 40 if(step == n+1) 41 42 if(mark[a[n]+a[1]] == false) // 判断最后一个数与第一个数 43 return; 44 for(int i = 1; i < n; ++i) 45 printf("%d ", a[i]); 46 printf("%d\n", a[n]); 47 48 49 50 for(int i = 2; i <= n; ++i) 51 52 if(!vis[i]) 53 54 a[step] = i; 55 vis[i] = 1; 56 dfs(step+1); 57 vis[i] = 0; 58 59 60 61 62 63 64 int main() 65 66 init(); 67 int cas = 1; 68 a[1] = 1; 69 while(scanf("%d", &n) != EOF) 70 71 printf("Case %d:\n", cas++); 72 memset(vis, 0, sizeof(vis)); 73 dfs(2); 74 printf("\n"); 75 76 77 78 return 0; 79
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