HDU1016：Prime Ring Problem（dfs）

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 52756    Accepted Submission(s): 23320

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

68

 

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

 

Source

Asia 1996, Shanghai (Mainland China)

题意：给n个数（1~n），要求组成一个环，要求相邻两个元素之和是素数，字典序输出所有组合。

心情差就想做水题。

# include <iostream># include <cstdio># include <cstring>using namespace std;int n, a[23], p[45], vis[23];void dfs(int pos){    if(pos == n+1)    {        for(int i=1; i<=n; ++i)            printf("%d%c",a[i],i==n?'\n':' ');        return;    }    for(int i=2; i<=n; ++i)    {        if(vis[i]||p[i+a[pos-1]]) continue;        if(pos == n && p[i+a[1]]) continue;        a[pos] = i;        vis[i] = 1;        dfs(pos+1);        vis[i]=0;    }}int main(){    int cas=1;    for(int i=2; i<=40; ++i)        if(p[i] == 0)            for(int j=i+i; j<=40; j+=i)                p[j] = 1;    while(~scanf("%d",&n))    {        a[1] = 1;        printf("Case %d:\n",cas++);        dfs(2);        puts("");    }    return 0;}