hdu1016 Prime Ring Problem(DFS)

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29755    Accepted Submission(s): 13268

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

68

 

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

 

Source

Asia 1996, Shanghai (Mainland China)

解题报告：因为蓝桥杯快到了，所以把以前做过的搜索题再重新做一遍。素数环，用过的数字标记就可以了。

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <cmath>using  namespace std;int arr[21], vis[21], n;bool isPrime(int n) {//n > 2if(n % 2 == 0) return 0;for (int i = 2; i <= sqrt(n); i ++) {if(n % i == 0) {return 0;}}return 1;}void dfs(int k) {if(n == k) {if(!isPrime(1 + arr[n-1])) return ;for (int i = 0; i < n-1; i ++) {printf("%d ", arr[i]);}printf("%d\n", arr[n-1]);return ;}for (int i = 2; i <= n; i ++) {if(!vis[i] && isPrime(arr[k-1] + i)) {arr[k] = i;vis[i] = 1;dfs(k+1);vis[i] = 0;}}}int main() {int cnt = 1;;while (scanf("%d", &n)!= EOF) {memset(vis, 0, sizeof(vis));vis[1] = 1;arr[0] = 1;printf("Case %d:\n", cnt ++);dfs(1);printf("\n");}return 0;}