hduoj1016,Prime Ring Problem，dfs题，重点

Prime Ring Problem

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

分析：

深搜题，用一个数组存储一个数是否被用过。并且由于题目特点判断质数不必用通常的方法，省下一点时间。

code:

#include<iostream>#include<cstdio>#include<cmath>#define MAX_SIZE 20using namespace std;int a[MAX_SIZE+5],n;int use[MAX_SIZE+5];const int prime[11]={3,5,7,11,13,17,19,23,29,31,37};bool belong(int n){    for(int i=0;i<11;i++)        if(prime[i]==n) return true;    return false;}void dfs(int r){    int i;    if(r==n&&belong(1+a[r-1]))//判断第一个数与最后一个数是否和为质数    {        printf("%d",a[0]);        for(i=1;i<n;i++)            printf(" %d",a[i]);        printf("\n");        return;    }    for(i=2;i<=n;i++)        if(use[i]==0&&belong(i+a[r-1]))        {            use[i]=1;            a[r]=i;            dfs(r+1);            use[i]=0;        }    return;}int main(){    int time=0;    while(scanf("%d",&n)==1)    {        time++;        memset(use,0,sizeof(use));        a[0]=1;        use[0]=1;        printf("Case %d:\n",time);        if(n%2==0) dfs(1);        printf("\n");    }    return 0;}