HDU(1016)搜索Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42265    Accepted Submission(s): 18729

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.
深搜算法：
找出一个环，使其临近的两数之和为质数，并按字典数排序：
从二开始搜索，把符合的数存在数组中，并标记已访问，注意细节
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

 

Source

Asia 1996, Shanghai (Mainland China)

#include<stdio.h>#include<cstring.h>#include<iostream>using namespace std;int v[25],num[25];int n;int prime[50]={0,1,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,0};void dfs(int m){if(m==n&&prime[num[0]+num[m-1]])//判断输出条件{for(int i=0;i<n-1;i++)printf("%d ",num[i]);printf("%d\n",num[n-1]);}elsefor(int i=2;i<=n;i++)//从2开始搜索{if(!v[i]&&prime[i+num[m-1]]){num[m++]=i;v[i]=1;//标记已搜索dfs(m);v[i]=0;//清除标记m--;}}}int main(){memset(v,0,sizeof(v));int t=1;while(~scanf("%d",&n)){printf("Case %d:\n",t);t++;num[0]=1;dfs(1);printf("\n");}}