HDOJ 1016 Prime Ring Problem（素数环 - 深搜）

题目

Prime Ring Problem

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 38   Accepted Submission(s) : 23

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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

68

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

 

 

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=1016

 

思路

  因为是做深度搜索的专题练习，所以大概思路并不难想，但是在dfs（）函数里却不知道以什么为结束条件，看了代码后，修改了一些地方，比如用save数组保存，用num计数，判断结束条件。

 

代码

#include<iostream>using namespace std;const int maxn = 20 +10;int used[maxn];int a[maxn];int save[maxn];int num;//记录当前找到第几个球int n;void dfs(int x){if(num == n-1 && a[x+1] == 0)//递归的结束条件 已经找到了n-1球，并且最后一个数跟1的和也是素数就输出{cout<<"1";for(int i=0; i<n-1; i++){cout<<" "<<save[i];}cout<<endl;return ;}for(int j=2; j<=n; j++){if(a[j + x] == 0 && used[j] == 0){save[num++] = j;//将j保存起来used[j] = 1;dfs(j);used[j] = 0;num--;//}}return;}int main(){int i,j;memset(a, 0, sizeof(a));a[0] = 1;a[1] = 1;for(i=2; i<maxn ;i++){if(a[i] != 0){continue;}for(j=i+i; j<maxn ;j=j+i){a[j] = 1;}}int count = 0;while(cin>>n){memset(used, 0, sizeof(used));count++;num = 0;cout<<"Case "<<count<<":"<<endl;used[1] = 1;if(n == 1)//没考虑到这种情况{cout<<"1"<<endl;}else dfs(1);cout<<endl;}return 0;}