hduoj1016,Prime Ring Problem，dfs题，重点

最新推荐文章于 2019-01-29 21:58:59 发布

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本文介绍了一个名为 PrimeRing 的问题，该问题要求在由 n 个节点组成的环中填入从 1 到 n 的自然数，使得任意相邻两节点上的数字之和均为质数。文章提供了一个 C++ 实现的解决方案，通过深度优先搜索算法来找出所有符合条件的填充方案。

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 20
using 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;
}