质数环-DFS

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22702    Accepted Submission(s): 10108


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

　　

#include<iostream>
#define MAX 25
using namespace std;
int a[MAX], b[MAX], n, t;
bool visit[MAX], prime[MAX*2];
int gcd(int x, int y){
    if (y == 0)return x;
    return gcd(y, x%y);
}
void dfs(int i, int pos){
    if (pos == n && prime[b[pos]+1]){
        for (int i = 1; i < pos; i++){
            cout << b[i] << " ";
        }
        cout << b[pos] << endl;
        return;
    }
    for (int j = 2; j <= n; j++){
        if (!visit[j] && prime[b[i] + a[j]]){
            visit[j] = true; b[i + 1] = a[j];
            dfs(i + 1, pos + 1);
            visit[j] = false;
        }
    }
    return;
}
int main()
{
    t = 0;
    while (cin >> n){
        for (int i = 1; i <= n; i++){
            a[i] = i;
        }
        fill(visit, visit + MAX, false);
        fill(prime, prime + MAX * 2, true);
        for (int i = 2; i < MAX * 2; i++){
            if (prime[i]){
                for (int j = i * 2; j < MAX * 2; j += i){
                    prime[j] = false;
                }
            }
        }
        cout << "Case " << ++t << ":" << endl;
        b[1] = 1; visit[1] = true;
        dfs(1, 1);
        cout << endl;
    }
    return 0;
}