HDUOJ1016 Prime Ring Problem

最新推荐文章于 2022-02-25 20:00:37 发布

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最新推荐文章于 2022-02-25 20:00:37 发布

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文章标签： J# C C++ C#

本文详细解析了素数环问题的解决方法，采用深度优先搜索（DFS）算法完成求解。通过递归的方式，实现对环中每个节点的遍历，并确保相邻节点之和为素数。文章提供了完整的代码实现及运行示例。

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Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8087Accepted Submission(s): 3620

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

【解题思路】素数环是典型的DFS题目，DFS函数的参数为(int c, int cnt)，

c代表遍历时所在环上节点的数，cnt代表结点个数。当节点个数位n时并且环末节点

和首节点相加也为素数则打印环。

根据本题总结dfs的主要步骤和方法：

1．找搜索出口条件，即搜索到最深层时的操作

2． Dfs采用递归搜索，在遍历后返回时标记已经遍历的节点

3．注意入口条件的满足

4．画出递归搜索图查看函数是否满足要求

5. 注意回溯的条件，适当的时候需要剪枝

//============================================================================ // Name : HDUOJ1016.cpp // Author : Weisi Shi // Version : // Copyright : Weisi Shi // Description : Hello World in C++, Ansi-style //============================================================================ #include <iostream> using namespace std; int n; bool p[41]; int circle[21]; bool visit[21]; void prime() { int i, j; memset(p, true, sizeof(p)); for (i = 2; i < 41; ++i) { for (j = 2; i * j < 41; ++j) { p[i * j] = false; } } } void dfs(int c, int cnt) { int i, j; if (cnt == n && p[circle[1] + circle[n]]) { for (i = 1; i < n; ++i) cout << circle[i] << " "; cout << circle[n] << endl; } for (j = 1, i = c + j; j <= n; i++, j++) { if (p[i] && !visit[j]) { circle[cnt + 1] = j; visit[j] = true; dfs(j, cnt + 1); visit[j] = false;//回溯 } } } int main() { int c = 1; prime(); while (cin >> n) { memset(visit, false, sizeof(visit)); cout << "Case " << c++ << ":" << endl; visit[1] = true; circle[1] = 1; dfs(1, 1); cout << endl; } return 0; }