HDU 1016: Prime Ring Problem

题目描述

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.

输入

n (0 < n < 20).

输出

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.

样例输入

6

8

样例输出

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

分析：dfs深搜，满足条件就加入num数组，并把加入的数置零。

因为数据范围在20以内，素数加和必定小于40，可以放个素数表～

#include <iostream>
#include <string>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <deque>
#include <map>
#define range(i,a,b) for(int i=a;i<=b;++i)
#define LL long long
#define rerange(i,a,b) for(int i=a;i>=b;--i)
#define fill(arr,tmp) memset(arr,tmp,sizeof(arr))
using namespace std;
int prime[15]={2,3,5,7,11,13,17,19,23,29,31,37};
int n,c,num[25],allnum[55];
void init() {
    c=0;
}
bool isprime(int num){
    range(i,0,11)if(num==prime[i])return true;
    return false;
}
bool dfs(int pre,int pos,int flag){
    if(!isprime(pre+pos))return false;
    num[flag]=pos;
    if(flag==n&&isprime(pos+1)){
        range(i,1,n)cout<<num[i]<<(i==n?'\n':' ');
        return true;
    }
    allnum[pos]=0;
    range(i,2,n)if(allnum[i]&&dfs(pos,i,flag+1))break;
    allnum[pos]=1;
    return false;
}
void solve(){
    while(cin>>n){
        range(i,1,n)allnum[i]=i;
        num[1]=1;
        cout<<"Case "<<++c<<":"<<endl;
        if(n==1){cout<<1<<endl;continue;}
        range(i,2,n)dfs(1,i,2);
        cout<<endl;
    }
}
int main() {
    init();
    solve();
    return 0;
}

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