HDU 1016（搜索题，DFS）

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5720 Accepted Submission(s): 2553


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此题纠结了我一个晚上。。。太悲剧了~ 一道简单的DFS题而已！
#include <iostream>
#include <math.h>
using namespace std;
#define N 20

int map

,mark
;
int res
,p
;
int n,ct;

bool is_prime(int u)
{
if(u==0 || u==1) return false;
if(u==2) return true;
if(u%2==0) return false;
for(int i=3;i<=sqrt((double)u);i+=2)
if(u%i==0)
return false;
return true;
}

void init()
{
int i,j;
for(i=1;i<N;i++)
{
for(j=i+1;j<N;j++)
{
if(is_prime(i+j))
map[i][j]=map[j][i]=1;
}
}
}

void output(int t)//输出例程
{
int k=0;
while(t)//通过对路径的分析获得结果
{
res[k++]=t;
t=p[t];
}
for(int j=n-1;j>=0;j--)//输出出来
{
if(j!=n-1)
putchar(' ');
printf("%d",res[j]);
}
putchar('/n');

}

void DFS(int x,int count)
{
mark[x]=1;
if(count==n && map[x][1]==1)
{
output(x);
}
for (int i=1;i<=n;i++)
{
if(map[x][i]==1 && mark[i]==0)
{
p[i]=x;
DFS(i,count+1);
mark[i]=0;
}
}
}

int main()
{
init();
while (scanf("%d",&n)!=EOF)
{
ct++;
printf("Case %d:/n",ct);
DFS(1,1);
printf("/n");
}
return 0;
}