HDU1035 Prime Ring Problem

                                                   Prime Ring Problem

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 14   Accepted Submission(s) : 10

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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)


解题思路：本题为典型一维深度优先搜索题。我学深搜时模仿的第一个代码，不知道怎么码解说了。。。。。。

#include<cstdio> #include<cstring> using namespace std; int n; int cil[22];    //已使用数据的存储 int num[22]; //数据使用情况0,1 int prime[40]={0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0};   // 素数列表，hass表 void dfs(int x) {     int i,j;     if(x==n&&prime[cil[0]+cil[x-1]])    //环已满，头尾能连接，符合要求     {         printf("1");         for(i=1;i<n;i++)             printf(" %d",cil[i]);         printf("\n");     }     else     {         for(i=2;i<=n;i++)         {             if(!num[i]&&prime[cil[x-1]+i])             {                 cil[x]=i;                 num[i]=1;                 dfs(x+1);                 num[i]=0;             }         }     } } int main() {     int i=1;     while(scanf("%d",&n)!=EOF)     {         memset(num,0,sizeof(num));         cil[0]=1;         printf("Case %d:\n",i++);         dfs(1);         printf("\n");     }     return 0; }