HDU 1016 Prime Ring Problem(DFS)

                          Prime Ring Problem

　　　　　　　　　　　　Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

　　　　　　　　　　　　　　　　　   Total Submission(s): 18636    Accepted Submission(s): 8350

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

 

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1016

 

题目大意：就是找出相邻两个数的和是素数的所有方案。

 

#include <stdio.h>
#include <string.h>
#include <math.h>

int N;
const int MAX_N = 25;
int arr[MAX_N];        //保存满足条件时各个数值
int visit[MAX_N];      //记录某个数是否访问过
int prime[MAX_N * 2];  //素数表

void IsPrime()         //判断是否是素数
{
    int i, k ;
    memset(prime, 0, sizeof(prime));
    for(int num = 2; num <= 45; num++)
    {
        k = sqrt((double)num);
        for(i = 2; i <= k; i++)
        {
            if(num % i == 0)
                break;
        }
        if(i > k)
            prime[num] = 1;
    }
}

void PrintResult()      //打印结果
{
    printf("1");
    for(int i = 2; i <= N; i++)
        printf(" %d", arr[i]);
    printf("\n");
}

void dfs(int curValue, int curCount) //curValue代表当前值，curCount代表当前数组下标
{
    if(curCount > N)return ;
    if(curCount == N && prime[curValue + 1]) //满足条件，输出答案
    {
        PrintResult();
        return ;
    }

    for(int i = 2; i <= N; i++)
    {
        if(visit[i] == 0 && prime[i + curValue])
        {
            visit[i] = 1;             //标记为已访问
            arr[curCount + 1] = i;
            dfs(i, curCount + 1);     //搜索下一个
            visit[i] = 0;             //回溯时恢复原来状态
        }
    }
}

int main()
{
    IsPrime();
    int nCase = 0;
    while(scanf("%d", &N) != EOF)
    {
        nCase++;
        memset(visit, 0, sizeof(visit));
        arr[1] = 1;
        visit[1] = 1;

        printf("Case %d:\n", nCase);
        dfs(1, 1);
        printf("\n");
    }
    return 0;
}