hdu(1016)

标签：

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33039    Accepted Submission(s): 14624

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

//这题的主要思路：直接利用dfs搜索满足头为1 两点之间的和为素数

#include <stdio.h>
int a[21],n,cnt;
int vis[21];
int cnt1=0;
int k[21];
int prime(int n)  //判断素数
{
    if(n==1) return 0;
    for(int i=2;i*i<=n;i++)
    {
        if(n%i==0)
            return 0;
    }
    return 1;
}

void dfs(int ini)
{
    if(cnt1==n&&prime(k[n-1]+1)&&k[0]==1)  //头为1 头尾相加为素数 长度为n
    {
        printf("%d",k[0]);
        for(int i=1;i<n;i++)
            printf(" %d",k[i]);
        printf("\n");
    }
    else
    {
        for(int i=0;i<n;i++)
        {
            if(!vis[i]&&prime(a[i]+a[ini]))
            {
                vis[i]=1;
                k[cnt1++]=a[i];
                dfs(i);
                cnt1--;  //回溯
                vis[i]=0;
            }
        }
    }

}

int main()
{
    cnt=1;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
        {
            a[i]=i+1;
            vis[i]=0;
        }
        printf("Case %d:\n",cnt++);
        dfs(0);
        printf("\n");
    }
    return 0;
}

hdu(1016)

标签：

原文地址：http://blog.csdn.net/a73265/article/details/46835249