HDU 1016 Prime Ring Problem(DFS入门)

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 54078    Accepted Submission(s): 23943

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

 DFS入门：注意结束条件

#include <stdio.h>#include<string.h>int n;int a[100],visit[100];int prime(int x){int i;for(i=2;i*i<=x;i++)if(x%i==0)return 0;return 1;}void dfs(int num){int i;if(num==n&&prime(a[num-1]+1)){for(i=0;i<n-1;i++)printf("%d ",a[i]);printf("%d\n",a[num-1]);}else{for(i=2;i<=n;i++){if(visit[i]==0){if(prime(a[num-1]+i)){visit[i]=-1;a[num++]=i;dfs(num);visit[i]=0;num--;}}}}}int main(int argc, char *argv[]){int flag=1;while(scanf("%d",&n)!=EOF){memset(visit,0,sizeof(visit));a[0]=1;printf("Case %d:\n",flag++);dfs(1);printf("\n");}return 0;}

2.0 版本

#include <stdio.h>#include<algorithm>#include<string>#include<set>using namespace std;int n;int visit[300];int a[300];bool prime(int n){int i;for(i=2;i*i<=n;i++){if(n%i==0)return false;}return true;}void dfs(int num) {int i,j;if(num==n&&prime(a[num-1]+1)){for(i=0;i<n-1;i++)printf("%d ",a[i]);printf("%d\n",a[n-1]);}else{for(i=2;i<=n;i++)//找到了进行下一层循环 找不到后退 向后递归时visit[i]自动变为 0  {if(prime(i+a[num-1])&&visit[i]==0){visit[i]=1;a[num]=i;dfs(num+1);visit[i]=0;}}}}int main(int argc, char *argv[]){int flag=1;while(scanf("%d",&n)!=EOF){memset(visit,0,sizeof(visit));a[0]=1;printf("Case %d:\n",flag++);dfs(1); printf("\n");}return 0;}