练习二 1020 Prime Ring Problem

题目：

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.<br><br>Note: the number of first circle should always be 1.<br><br><img src=../../data/images/1016-1.gif><br>

Input

n (0 < n < 20).<br>

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.<br><br>You are to write a program that completes above process.<br><br>Print a blank line after each case.<br>

Sample Input

6<br>8<br>

Sample Output

Case 1:<br>1 4 3 2 5 6<br>1 6 5 2 3 4<br><br>Case 2:<br>1 2 3 8 5 6 7 4<br>1 2 5 8 3 4 7 6<br>1 4 7 6 5 8 3 2<br>1 6 7 4 3 8 5 2<br>
题目大意：输入一个 n 找出1~n的组合，使得相邻两个数之和为素数
解题思路：这道题非常类似于N皇后问题，可以使用的深度优先搜索方法来解决，虽然说打印要求是按照顺时针和逆时针顺序打印，其实按照从小到大的搜索顺序搜索后的结果就是符合输出顺序的，而且由于是20以内的数字，所以判断质数的方法是直接打表后一个简单的循环判断一下是否为质数并标记某个数字是否使用过，最终可求出结果。
感想：深搜问题，还好老师讲过类似题目，要不然肯定很难做出来。
代码：
#include<iostream>#include<cstring>#include<cstdio>#include<cstdlib>#include<iomanip>using namespace std;int num[21],mark[21],n;int prime_num[12]={2,3,5,7,11,13,17,19,23,29,31,37}; //判断是否是质数，是返回1，不是返回0int 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++)    cout<<num[i]<<" ";    cout<<num[n]<<" ";}int dfs(int pre,int post,int flag){    if(!is_prime(pre+post)) return 0; //如果不符合，直接返回    num[flag]=post;    if(flag==n&&is_prime(post+1))    {        print_num();        printf("\n");        return 1;    }    mark[post]=0; //使用过了这个数字就标记为0    for(int i=2;i<=n;i++)    if(mark[i]!=0&&dfs(post,i,flag+1)) break;    mark[post]=1;//标记位恢复原状    return 0;}int main(){    int count,i;    count=1;    while(cin>>n)    {        for(int i=1;i<=n;i++)        mark[i]=i;        num[1]=1;        cout<<"Case "<<count++<<":"<<endl;        if(n==1)cout<<"1"<<endl;        for(i=2;i<=n;i++)        dfs(1,i,2);        cout<<endl;    }    return 0;}