HDU1016-Prime Ring Problem（DFS）

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

 

分析：

题意：输入正整数n，把整数1,2,3,…,n组成一个环，使得相邻两个整数之和均为素

数。输出时从整数1开始逆时针排列。同一个环应恰好输出一次。

素数环问题的程序实际上主要由求素数和整数1,2,3,…,n的排列构成。

运用DFS模型来回溯、其处理效率提高、看代码就能明白整个处理过程。

#include<iostream>#include<string.h>#include<stdio.h>#include<ctype.h>#include<algorithm>#include<stack>#include<queue>#include<set>#include<math.h>#include<vector>#include<map>#include<deque>#include<list>using namespace std;bool is_prime(int x)   //判数一个整数x是否是一个素数{    for(int i = 2; i*i <= x; i++)    {        if(x % i == 0)            return 0;    }    return 1;}int n, A[50], isp[50], vis[50];int w=1;void dfs(int p){    if(p == n && isp[A[0]+A[n-1]])  //递归边界，别忘测试第一个数和最后一个数    {        for(int i = 0; i < n-1; i++)            printf("%d ", A[i]);        printf("%d\n",A[n-1]);    }    else        for(int i = 2; i <= n; i++)  //尝试放置每个数i            if(!vis[i] && isp[i+A[p-1]])            {       //如果i没有用过，并且与前一个数之和为素数                A[p] = i;                vis[i] = 1; //设置标记                dfs(p+1);                vis[i] = 0; //清除标记            }}int main(){    while(scanf("%d", &n)!=EOF)    {        for(int i = 2; i <= n*2; i++)            isp[i] = is_prime(i);        memset(vis, 0, sizeof(vis));        A[0] = 1;        printf("Case %d:\n",w++);        dfs(1);        printf("\n");    }    return 0;}