hdu 1016 Prime Ring Problem (DFS)

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21703    Accepted Submission(s): 9682

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

 

 

Source

Asia 1996, Shanghai (Mainland China)

 

 

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//250MS    232K    1064 B    G++
/*

    题意：
        给出一个数n，求数1到n组成的环，环相邻的两个数和为素数。输出符合条件的环。
        
    DFS:
        这道题我开始看了几次都没动手，怕功力不够写不出来。后来最多了再回来做，
    发现也算是一道基础题而已。
        因为n比较小，我这里先预处理符合条件的一对数的情况，保存在g中。然后直接深搜，
    再用pre数组保存没个数的前驱，最后将符合条件的pirnt出来。 

*/
#include<stdio.h>
#include<string.h>
int g[25][25]; //预处理 
int vis[25];
int pre[25]; //前一个数 
int n;
int judge(int x)
{
    for(int i=2;i*i<=x;i++)
        if(x%i==0) return 0;
    return 1;
}
void init()
{
    memset(g,0,sizeof(g));
    for(int i=1;i<25;i++)
        for(int j=i+1;j<25;j++)
            if(judge(i+j)) g[i][j]=g[j][i]=1;
}
void print(int x)
{
    int ans[25],i=1;
    ans[0]=x;
    while(pre[x]!=-1){
        ans[i++]=pre[x];
        x=pre[x];
    }
    for(int i=n-1;i>0;i--) printf("%d ",ans[i]);
    printf("%d\n",ans[0]);
}
void dfs(int x,int cnt)
{
    if(cnt==n && g[x][1]) print(x); //第一个和最后一个的和也是素数 
    for(int i=2;i<=n;i++){
        if(!vis[i] && g[x][i]){
            pre[i]=x;
            vis[i]=1;
            dfs(i,cnt+1);
            vis[i]=0;
        }
    }
}
int main(void)
{
    int k=1;
    init();
    while(scanf("%d",&n)!=EOF)
    {
        printf("Case %d:\n",k++);
        memset(vis,0,sizeof(vis));
        pre[1]=-1;
        vis[1]=1;
        dfs(1,1);
        printf("\n");
    }
    return 0;
}