HDU-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): 46985 Accepted Submission(s): 20746
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
思路:
简单的DFS,做过突然想温习一下,旧题新写
代码:
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<cmath> #include<string> using namespace std; const int MAXN=21; bool flag[MAXN]; int ans[MAXN]; int total=0; int t=1; int N; bool check(int x, int y) { x+=y; int i=2; while(i<=sqrt(x)&&x%i) { i++; } if(i>sqrt(x)) return true; return false; } void print() { printf("%d",ans[1]); for(int i=2;i<=N;i++) { printf(" %d",ans[i]); } printf("\n"); } void dfs(int num) { for(int i=2;i<=N;i++) { if(check(ans[num-1],i)&&!flag[i]) { ans[num]=i; flag[i]=true; if(num==N) { if(check(ans[num],ans[1])) print(); } else { dfs(num+1); } flag[i]=false; } } } int main() { while(scanf("%d",&N)!=EOF) { memset(flag,false,sizeof(flag)); printf("Case %d:\n",t); ans[1]=1; dfs(2); printf("\n"); t++; } return 0; }