HDU - 1016 Prime Ring Problem (DFS) (day_8_F)

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 70220 Accepted Submission(s): 30012

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)

Recommend
JGShining

题目简述：

输入n，输出某种排序使得，从1~n的数字与任意相邻一个数字的和为素数

题目分析：

DFS遍历所有可能的排序，输出正确的

代码实现：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<stack>
#include<cstdlib>
#include<cmath>
#include<queue>
//#include <bits/stdc++.h>
using namespace std;

const int INF = 0x3f3f3f3f;
#define pf          printf
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define ms(i,j) memset(i,j,sizeof(i))

bool mark[22],prime[44]={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};
int n,len,k=1;
int a[21];
void dfs(int num)
{
    if(num==n&&prime[a[n]+1])
    {
        for(int i=1;i<n;i++)
            pf("%d ",a[i]);
        pf("%d\n",a[n]);
        return;
    }
    for(int i=2;i<=n;i++)
        if(!mark[i]&&prime[a[num]+i])
        {
            mark[i]=1;
            a[num+1]=i;
            dfs(num+1);
            mark[i]=0;
        }
}
int main()
{
    a[1]=1;
    while(~sf(n))
    {
        ms(mark,0);
        pf("Case %d:\n",k++);
        dfs(1);
        pf("\n");
    }
}