HDOJ Prime Ring Problem (深度优先搜索)
Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11488 Accepted Submission(s): 5194
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.
![](http://acm.hdu.edu.cn/data/images/1016-1.gif)
Note: the number of first circle should always be 1.
![](http://acm.hdu.edu.cn/data/images/1016-1.gif)
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.
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
#include <iostream> #include <cstring> using namespace std; int n,j,k=1,visit[20],a[20]={1}; 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}; void DFS(int m) { if(m==n && prime[1+a[n-1]]) { for(int i=0;i<n-1;i++) cout<<a[i]<<" "; cout<<a[n-1]<<endl; } else { for(int i=2;i<=n;i++) { if(prime[i+a[m-1]] && !visit[i]) { a[m]=i; visit[i]=1; DFS(m+1); visit[i]=0; } } } } int main() { while(cin>>n) { memset(visit,0,sizeof(visit)); cout<<"Case "<<k++<<":"<<endl; DFS(1); cout<<endl; } return 0; }