HDOJ1016（搜索DFS）

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38643    Accepted Submission(s): 17073

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

 
此题目是dfs问题，要回回溯

#include<iostream>#include<cstring>#include<cstdio>#include<cmath>using namespace std;int n;int a[21];//存放结果int sign[21];//用来标记int isprime[38]={0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1};void dfs(int m){    //结束条件    if(m==n&&isprime[a[n-1]+a[0]])//因为是环，判断最后一位和第一位的和    {        cout << a[0];        for(int i=1;i<n;i++)            cout << " " << a[i];        cout << endl;    }    else    {        for(int i=2;i<=n;i++)        {            if(!sign[i]&&isprime[i+a[m-1]])            {                a[m]=i;                sign[i]=1;//已经用过，标记为1                dfs(m+1);//递归                sign[i]=0;//如果递归回来没有用到这个数，重新标记为可用的            }        }    }}int main(){    int count=0;    a[0]=1;    while(~scanf("%d",&n))    {        memset(sign,0,sizeof(sign));        count++;        cout << "Case " << count <<":" << endl;        dfs(1);        cout << endl;    }    return 0;}