hdu1016(dfs实现环状全排序)

Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 39348 Accepted Submission(s): 17338

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(顺时针方向的) and anticlockwisely(逆时针的). The order of numbers must satisfy the above requirements. Print solutions in lexicographical(辞典编纂的) 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

题解：第一个数从2到n开始往下找，标记此数，然后继续第二个数从2到n开始往下找，标记此数，找到了第n个数的时候把它打印出来，然后往回退一个，继续往下找，以此类推，直到找完所有的情况为止。

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>using namespace std;//int prime[42]={0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1};int vis[21],a[21],n;bool prime(int n){    int k,i;    if(n<2)return false;    for(i=2;i<=(int)sqrt(n);i++){        if(n%i==0)return false;    }       return true;}void dfs(int k){    int i;    if(k==n&&prime(a[0]+a[n-1])){        for(i=0;i<n-1;i++)            printf("%d ",a[i]);        printf("%d\n",a[n-1]);    }    else{        for(i=2;i<=n;i++){            if(!vis[i]){                if(prime(a[k-1]+i)){                    vis[i]=1;                    a[k++]=i;                    dfs(k);                    vis[i]=0;                    k--;                }            }        }           }}int main(){    int i;    i=1;    while(scanf("%d",&n)!=EOF){        printf("Case %d:\n",i);        memset(vis,0,sizeof(vis));        a[0]=1;        dfs(1);        i++;            printf("\n");    }       return 0;}