HDU 1016--DFS第一炮

最新推荐文章于 2025-09-06 09:33:29 发布

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最新推荐文章于 2025-09-06 09:33:29 发布

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原文链接：http://www.cnblogs.com/hankers/archive/2012/01/25/2329339.html

本文介绍如何使用深度搜索算法解决素数环问题，即在一定范围内将自然数排列成圆环，使得相邻两数之和为素数。通过构建素性数组并递归搜索符合条件的序列，实现算法求解。

HDU 1016

--DFS第一炮

题目：

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10626 Accepted Submission(s): 4800

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

Source
Asia 1996, Shanghai (Mainland China)

Recommend
JGShining

分析：

该题讲述了要求一个素数环，使得相邻的两个数之和为素数，求其各种素数环。由于它为层层递进，故想到深度搜索DFS，又由于0<n<20,则简单地设置了一个素性数组。

代码:

 1 #include<stdio.h>
 2 #include<string.h>
 3 
 4 int prime[39]={0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0};//判定1-38素性数组
 5 int ans[20],visit[20],num;
 6 
 7 void dfs(int n)
 8 {
 9      int i;
10      if(n==num&&prime[ans[0]+ans[n-1]]==1)
11      {
12           for(i=0;i<n-1;i++)
13               printf("%d ",ans[i]);
14           printf("%d\n",ans[n-1]);
15      }
16      else
17      {
18           for(i=2;i<=num;i++)
19           {
20               if(!visit[i]&&prime[ans[n-1]+i]==1)
21               {
22                   visit[i]=1;
23                   ans[n]=i;
24                   dfs(n+1);
25                   visit[i]=0;
26               }
27           }
28      }
29 }
30 
31 int main()
32 {
33     int i=1;
34     ans[0]=1;
35     while(scanf("%d",&num)!=EOF)
36     {
37         memset(visit,0,sizeof(visit));
38         printf("Case %d:\n",i++);
39         dfs(1);
40         printf("\n");
41     }
42     return 0;
43 }