杭电 1016 Prime Ring Problem【DFS】

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34175    Accepted Submission(s): 15122

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

 

解题思路：

1.这题采用了查表法，用prime数组代替素数判断的方式，节省时间。

2.在dfs函数中新增了判断n的奇偶性，因为如果n为奇数，必定多出来一个奇数，根据奇偶性原理，奇+奇=偶，凡是偶数都不是素数，必错。

3.在judge函数中，通过循环剔除重复的成分，否则可能有重复因子(不信可以注释掉后试试)。

4.在dfs函数中没有包含首尾元素的判断，由于不可能为重，所以直接判断加和是否为素数即可。

备注：

dfs函数的t为数组角标，i为对应的数。

#include<stdio.h>#include<string.h>int a[20],n;int prime[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,};bool judge(int last,int x){int i;if(!((a[last]+x)&1)) return 0;if(!prime[a[last]+x]) return 0;for(i=0;i<=last;i++){if(a[i]==x) return 0;}return 1;}void dfs(int t){int i;if(n&1) return;if(t==n){if(prime[a[0]+a[n-1]]){printf("%d",a[0]);for(i=1;i<n;i++){printf(" %d",a[i]);}printf("\n");}}else{for(i=2;i<=n;i++){a[t]=i;if(judge(t-1,i)){dfs(t+1);}}}}int main(){int k=1;while(~scanf("%d",&n)){memset(a,0,sizeof(a));a[0]=1;printf("Case %d:\n",k++);dfs(1);printf("\n");}return 0;}