HDU 1016.Prime Ring Problem-素数环，相邻两数和为素数-DFS

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 67252    Accepted Submission(s): 28829

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)

 

 

 

 

 

代码:

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int prime[]={2,3,5,7,11,13,17,19,23,29,31,37};
bool isprime[40];
bool used[21];
int num[21],n;
bool dfs(int count,int cur){ //深度优先搜索
    num[count]=cur;
    if(count==n-1){
        if(!isprime[cur+1])return false;
        for(int i=0;i<n-1;i++)
            printf("%d ",num[i]);
        printf("%d
",num[n-1]);
        return false;
    }
    for(int i=2;i<=n;i++){
        if(used[i])continue;
        used[i]=true;
        if(isprime[cur+i]&&dfs(count+1,i))
            return true;
        used[i]=false;
    }
    return false;
}
int main(){
    memset(isprime,false,sizeof(isprime));
    for(int i=0;i<12;i++)
        isprime[prime[i]]=true;//存在素数isprime数组标记为1
    int Case=1;
    while(~scanf("%d",&n)){
        printf("Case %d:
",Case++);
        memset(used,false,sizeof(used));
        used[1]=true;
        dfs(0,1);
        printf("
");
    }
    return 0;
}