hdu1016 Prime Ring Problem【素數環問題（經典dfs）】

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 74594    Accepted Submission(s): 31690

 

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)

 

題意大體就是說要找到由1~n組成的環，且相鄰的數兩兩相加都要是素數，輸出全部知足條件的序列。

AC代碼以下：

#include<stdio.h>#include<string.h>int f[100]={0};int ans[21],v[21],n;void dfs(int i){    int m;    if(i==n&&f[ans[i-1]+ans[0]]==0) //找到8個知足條件的數時輸出    {        for(m=0;m<n-1;m++)            printf("%d ",ans[m]);        printf("%d\n",ans[n-1]);    }    else                          {        for(m=2;m<=n;m++)        //找未用過的數        {            if(v[m]==0)          //找到尚未用過的數            {                if(f[m+ans[i-1]]==0)  //若是知足相鄰數和爲素數                {                    v[m]=-1;     //標記已用過                    ans[i++]=m;  //將m放進數組                    dfs(i);      //找下一個數                    v[m]=0;      //接觸標記                    i--;         //回溯                }            }        }    }}int main(){    int i,j,cases=0;    //素數打表    for(i=2;i<100;i++)        if(f[i]==0)            for(j=i;i*j<100;j++)                f[i*j]=1;        while(scanf("%d",&n)!=EOF)    {        cases++;        ans[0]=1;        printf("Case %d:\n",cases);        memset(v,0,sizeof(v));        dfs(1);        printf("\n");    }    return 0;}--------------------- 做者：yongtaozheng 來源：CSDN 原文：https://blog.csdn.net/Twinkle_sone/article/details/95123292 版權聲明：本文爲博主原創文章，轉載請附上博文連接！