hdu 1016 Prime Ring Problem（深度優先搜索）

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14715    Accepted Submission(s): 6720

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

 

 

Source

Asia 1996, Shanghai (Mainland China)

分析：

（1）這道題非常類似於N皇后問題，使用的是深度優先搜索方法

（2）雖然說打印要求是按照順時針和逆時針順序打印，其實按照從小到大的搜索順序搜索后的結果就是符合輸出順序的。

（3）由於是20以內的數字，所以判斷質數的方法是直接打表后一個簡單的循環判斷一下是否為質數

（4）程序中mark數組是為了標記某個數字是否使用過了，num數組存儲的是數字鏈表的存儲順序。

#include <stdio.h>

int num[21],mark[21],n;
int prime_num[12] = {2,3,5,7,11,13,17,19,23,29,31,37};

//判斷是否是質數，是返回1，不是返回0
int is_prime(int a)
{
    for(int i = 0; i < 12;i++)
    if(a==prime_num[i])return 1;
    return 0;
}
void print_num()
{
    for(int i = 1; i < n;i++)
    printf("%d ",num[i]);
    printf("%d",num[n]);
}

int dfs(int pre,int post,int flag)
{
    //如果不符合，直接返回
    if(!is_prime(pre+post))
    return 0;
    num[flag] = post;
    if(flag==n&&is_prime(post+1))
    {
        print_num();
        printf("\n");
        return 1;
    }
    //使用過了這個數字就標記為0
    mark[post] = 0;
    for(int i = 2;i<=n;i++)
    if(mark[i]!=0 && dfs(post,i,flag+1))break;
    //標記位恢復原狀
    mark[post] = 1;
    return 0;
}

int main()
{
    int count;
    count = 1;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i = 1; i <= n; i++)
        mark[i] = i;
        num[1] = 1;
        printf("Case %d:\n",count++);
        if(n==1)printf("1\n");
        for(int i = 2;i<=n;i++)
        dfs(1,i,2);
        printf("\n");
    }
    return 0;
}