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
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
简单dfs dfs有点菜所以练了一下。此题有两个状态：一个是递归深度（即为数组下标），一个是此时数。dfs也做了不少了，总结一点，dfs实际上是对问题的状态空间的深度优先遍历，是对树的深度优先遍历的映射。和普通暴力的区别就是遍历的方式不同。而且因为有剪枝的存在，能够减低时间复杂度。
题解：
//1016
#include<iostream>
#include<cstring>
#include<string>
using namespace std;
int n;
int ans[22];
bool vis[22];
int isprime(int x)
{
int i;
if(x<2) return 0;
else if(x == 2) return 1;
if(x%2 == 0) return 0;
for(i = 3;i*i<=x;i+=2)
if(x%i == 0) return 0;
return 1;
}
void dfs(int dep,int now)//dep 深度 now 此时的数
{
if(vis[now]) return ;//跑过直接回溯
if(!isprime(now+ans[dep-1])) return;//满足题目中相邻的相加是素数
if(dep == n)//当深度是n时
{
if(!isprime(now+ans[1])) return ;//不能忘了最后还要和第一个比较一下
else {
ans[dep] = now;
cout<<1;
for(int i =2;i<=n;i++)
cout<<" "<<ans[i];
cout<<endl;
return ;
}
}
for(int i=2;i<=n;i++)//符合就递归，不符合就回溯
{
ans[dep] = now;
vis[now] = 1;
dfs(dep+1,i);
vis[now] = 0;
}
}
int main ()
{
int kase = 0;
while (cin >> n)
{
cout << "Case " << ++kase << ":" << endl;
memset(vis, 0, sizeof(vis));
ans[0] = 1;
dfs(1, 1);
cout << endl;
}
return 0 ;
}