hdu 1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19251    Accepted Submission(s): 8599

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)

 

Recommend

JGShining

解题思路：

素数打表，因为n最大是20，所以只要打到40

用DFS进行搜索，然后用vis进行标记，搜索过还需要回溯，为了查找下一组数据，本题数据比较水，不需要简直都可以过

最后还要注意格式：两个数之间要有空格，每组数据要有空行

#include<stdio.h>#include<cstring>using namespace std;int prime[40]={0,1,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0};//素数打表，因为n最大是20，所以只要打到40int vis[21],a[21],N;void dfs(int num){if(num==N&&prime[a[num-1]+a[0]]){for(int i=0;i<num-1;i++)printf("%d ",a[i]);printf("%d\n",a[num-1]);}else{for(int i=2;i<=N;i++){if(vis[i]==0){if(prime[i+a[num-1]]){vis[i]=1;a[num++]=i;dfs(num);vis[i]=0;num--;}}}}}int main(){int num=1;while(scanf("%d",&N)!=EOF){printf("Case %d:\n",num++);memset(vis,0,sizeof(vis));a[0]=1;dfs(1);printf("\n");}return 0;}