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

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

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0,1int 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;}