ZOJ 1457 E-Prime Ring Problem

https://vjudge.net/contest/67836#problem/E

 

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

 

时间复杂度：$O(n!)$

题解：dfs 要先判断 N ，如果 N 是奇数的情况下不可能构成素数环，而且如果不先排除会报超时

代码：

#include <bits/stdc++.h>
using namespace std;

int N;
int vis[30];
int a[30];

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

void dfs(int step) {
    if(step == N + 1 && prime(a[1] + a[N])) {
        for(int i = 1; i <= N; i ++) {
            printf("%d", a[i]);
            printf("%s", i != N ? " " : "\n");
        }
        return ;
    }
    for(int i = 2; i <= N; i ++) {
        if(vis[i] == 0) {
            if(prime(i + a[step - 1])) {
                vis[i] = 1;
                a[step] = i;
                dfs(step + 1);
                vis[i] = 0;
            }
        }
    }
    return ;
}

int main() {
    int cnt = 0;
    while(~scanf("%d", &N)) {
        memset(vis, 0, sizeof(vis));
        memset(a, 0, sizeof(a));
        printf("Case %d:\n", ++cnt);
        if(N % 2) {
            printf("\n");
            continue;
        }
        a[1] = 1;
        vis[1] = 1;
        dfs(2);
        printf("\n");
    }
    return 0;
}