HDOJ-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): 47217    Accepted Submission(s): 20859

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

 

代码：

#include <iostream>
#include <cstring>
using namespace std;
bool prime[38] = { 0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1 };
bool vis[22];
int n, ring[22];
void ans() {
    cout << ring[0];
    for (int i = 1; i < n; i++) {
        cout << ' ' << ring[i];
    }
    cout << '\n';
}
void dfs(int pos) {
    if (pos == n &&prime[ring[n - 1] + ring[0]]) {
        ans();
        return;
    }
    for (int i = 2; i <= n; i++) {
        if (!vis[i] && prime[i + ring[pos - 1]]) {
            vis[i] = true;
            ring[pos] = i;
            dfs(pos + 1);
            vis[i] = false;
        }
    }
}
int main()
{
    int case_num = 1;
    ring[0] = 1;
    while (cin >> n) {
        memset(vis, false, sizeof(vis));
        cout << "Case " << case_num << ":" << endl;
        case_num++;
        dfs(1);
        cout << endl;
    }
    return 0;
}