UVa 524 Prime Ring Problem（DFS , 回溯）

题意 把1到n这n个数以1为首位围成一圈 输出全部满足随意相邻两数之和均为素数的全部排列

直接枚举排列看是否符合肯定会超时的 n最大为16 利用回溯法 边生成边推断 就要快非常多了

#include<cstdio>
using namespace std;
const int N = 50;
int p
, vis
, a
, n;

int isPrime(int k)
{
for(int i = 2; i * i <= k; ++i)
if(k % i == 0) return 0;
return 1;
}

void dfs(int cur)
{
if(cur == n && p[a[n - 1] + 1])
{
printf("%d", a[0]);
for(int i = 1; i < n; ++i)
printf(" %d", a[i]);
printf("\n");
}

for(int i = 2; cur < n && i <= n; ++i)
{
if(!vis[i] && p[a[cur - 1] + i])
{
vis[i] = a[cur] = i;
dfs(cur + 1);
vis[i] = 0;
}
}
}

int main()
{
int cas = 0;
a[0] = 1;
for(int i = 2; i < N; ++i)
p[i] = isPrime(i);

while(~scanf("%d", &n))
{
if(cas) printf("\n");
printf("Case %d:\n", ++cas);
dfs(1);
}

return 0;
}

Prime Ring Problem

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers

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 <= 16)

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.

You are to write a program that completes above process.

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