HDU1016 Prime Ring Problem（DFS回溯）

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34609    Accepted Submission(s): 15327

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

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

 

这道题就是回溯暴力。 首先打出一个素数表， 然后DFS回溯判断即可。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int is_prime(int x)
{
    for(int i=; i*i<=x; i++)
    if(x%i==) return ;
    return ;
}

int n, A[];
bool isp[], vis[];
void dfs(int cur)
{
    if(cur==n&&isp[A[]+A[n-]])
    {
        for(int i=; i<n-; i++)
        {
            printf("%d ", A[i]);
        }
        printf("%d\n", A[n-]);
    }
    else for(int i=; i<=n; i++)
    if(!vis[i]&&isp[i+A[cur-]])
    {
        A[cur] = i;
        vis[i] = ;
        dfs(cur+);
        vis[i]=;
    }
}

int main()
{
    int kase = ;
    while(scanf("%d", &n)!=EOF)
    {
        printf("Case %d:\n", ++kase);
        for(int i=; i<=n*; i++)
        isp[i] = is_prime(i);
        memset(vis, , sizeof(vis));
        A[] = ;
        dfs();
        printf("\n");
    }
    return ;
}