HDU 1016 Prime Ring Problem【第一次DFS的邂逅】

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 34498 Accepted Submission(s): 15267



[align=left]Problem Description[/align]
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.



[align=left]Input[/align]
n (0 < n < 20).

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

6
8

[align=left]Sample Output[/align]

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

[align=left]Source[/align]
Asia 1996, Shanghai (Mainland China)

注意：

1：在fun函数里面的i不可以用作全局变量 因为DFS的递归（函数调用）会影响i的值 所以i只能在fun函数里面定义（PS：除非i在递归中没有数值的影响）

2：深搜的模板(自己编的)

void fun(int a)
{
if()/*剪枝的条件*/
{
;
}
if()/*非法情况的条件*/
{
;
}
if()/*最深层递归的跳出条件*/
{
;
}
if()/*继续递归条件*/
{
vis[i]=false;/*深搜如果不能返回前一个*/
fun(num);/*递归调用*/
vis[i]=true;/*返回递归的一个退出条件*/
}
}

代码如下：

#include<stdio.h>
#include<string.h>
#include<stdbool.h>
int n;
bool vis[25];
int num[25];
bool fun(int s)
{
for(int i=2; i*i<=s; i++)
{
if(s%i==0)
return false;
}
return true;
}
void Dfs(int s,int t)//s第t-1个数的值  t要搜索第t个数
{
if(t==n+1)//跳出的条件
{
if(fun(s+1)==true)
{
for(int i=1; i<n; i++)
{
printf("%d ",num[i]);
}
printf("%d\n",num
);
}
return ;
}
for(int i=1; i<=n; i++)//递归调用
{
if(vis[i]==false)//判断是否重复出现
{
if(fun(i+s)==true)
{
num[t]=i;
vis[i]=true;
Dfs(i,t+1);
vis[i]=false;//进入到最深层后，返回时用到
}
}
}
}
int main()
{
int Case=0;
while(~scanf("%d",&n))
{
printf("Case %d:\n",++Case);
memset(vis,0,sizeof(vis));
vis[1]=true;//确定这个数字使用过
num[1]=1;//输出的数字
Dfs(1,2);
printf("\n");
}
return 0;
}