杭电 HDU ACM 1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31900    Accepted Submission(s): 14108

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)

 

 

又一个深搜题目！并且这个素数环问题 参考了 人家的优化方案，值得好好吸收。搜索过程可以从实际问题中抽象而不是单纯的 ”矩阵给出“。奇偶性剪枝也要时刻注意。

#include
#include
#include
using namespace std;

int prime[41],visited[21],step,n;
listcnt;
void Prime()             // 素数打表判定
{
    memset(prime,0,sizeof(prime));
    for(int i=2; i<=41; i++)
    {
        if(!prime[i])
        {
            for(int j=i+i; j<=41; j+=i)
                prime[j]=1;
        }
    }
    prime[1]=1;
}

void dfs(int k)
{

    if(step==n&&!prime[k+1])  //某次搜索结束条件，也即如果出现某条满足题意搜索，输出即可
    {
        cout<<1;
        for(list::iterator it=cnt.begin(); it!=cnt.end(); it++)
            cout<<" "<<(*it);
        cout<>n)
    {
        cout<<"Case "<<++t<<":"<