hdoj1016-素数环递归

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题目描述

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.

image.png

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

1
2
6
8

Sample Output

1
2
3
4
5
6
7
8
9
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

思路分析

这题对我来说有些难，最主要是这个递归的过程有点难懂。还是要多看看代码才行，贴出来的代码还是能懂的，但是自己写却比较困难了。

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#include <cstdio>

int ans[22];
int visit[22];
int n;
int prime[] = {2,3,5,7,11,13,17,19,23,29,31,37,41};
bool judgePrime(int x){
    for(int i=0; i<13; i++){
        if(prime[i] == x) return true;
    }
    return false;
}
void DFS(int x){
    if(x == n){
        if(judgePrime(1+ans[n])){
            for(int i=1; i<=n; i++){
                if(i != 1) printf(" ");
                printf("%d", ans[i]);
            }
            printf("\n");
        } 
        return;
    }
    else{
        for(int i=2; i<=n; i++){
            if(!visit[i] && judgePrime(i + ans[x])){
                visit[i] = 1;
                ans[x+1] = i;
                DFS(x+1);
                visit[i] = 0; // 这里需要重置为0，以便于下次递归入口使用。
            }
        }
    }
}
int main(){
    int cas = 0;
    while(scanf("%d", &n) !=EOF){
        cas++;
        for(int i=0; i<22; i++) visit[i] = 0;
        ans[1] = 1;
        printf("Case %d:\n", cas);
        visit[1] = true;
        DFS(1);
        printf("\n");
    }
}