UVA - 524 Prime Ring Problem （经典回溯问题）

  Prime Ring Problem 

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 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 <= 16)

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.

You are to write a program that completes above process.

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

题目大意：输入正整数n，把1,2,3,……n组成一个环，使得相邻的两个整数之和均为素数。输出时从整数1开始逆时针排列，同一个换恰好输出一次。n<=16

解析：本来我是想用全排列，筛选出素数环的，但是无奈，时间效率比较低。于是参考了刘汝佳的题解敲了一遍，加深了对于回溯的认识。

#include <stdio.h>#include <math.h>#include <string.h>#include <algorithm>using namespace std;const int N = 50;int isp[N];int A[N];int vis[N];int n;int is_prime(int n) {for(int i = 2; i <= sqrt(n); i++) {if( n % i == 0) {return 0;}}return n;}void dfs(int cur) {if(cur == n && isp[A[0] + A[n-1]]) { //递归边界for(int i = 0; i < n; i++) {printf("%d", A[i]);if( i != n-1) {printf(" ");}}printf("\n");}else {for(int i = 2; i <= n; i++) {if( !vis[i] && isp[i+A[cur-1]] ) { //如果i没有用过，并且与前一个数之和为素数A[cur] = i;vis[i] = 1;dfs(cur+1);vis[i] = 0;}}}}int main() {int cas = 0;isp[0] = 0;isp[1] = 0;for(int i = 2; i < N; i++) {isp[i] = is_prime(i);}while( scanf("%d",&n) != EOF) {memset(vis,0,sizeof(vis));if(cas++) {printf("\n");}for(int i = 0; i < n; i++) {A[i] = i+1;}printf("Case %d:\n",cas);dfs(1);}return 0;}