HDOJ1016（搜索DFS）

最新推荐文章于 2018-02-28 13:41:25 发布

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本文详细阐述了如何通过深度优先搜索（DFS）算法解决PrimeRingProblem，即在一个圆环中放置从1到n的自然数，使得任意相邻两个数之和均为质数。文中提供了完整的代码实现，并解释了输入输出格式及解决方案的生成过程。

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Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38643 Accepted Submission(s): 17073

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

此题目是dfs问题，要回回溯

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
int n;
int a[21];//存放结果
int sign[21];//用来标记
int isprime[38]={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};
void dfs(int m)
{
    //结束条件
    if(m==n&&isprime[a[n-1]+a[0]])//因为是环，判断最后一位和第一位的和
    {
        cout << a[0];
        for(int i=1;i<n;i++)
            cout << " " << a[i];
        cout << endl;
    }
    else
    {
        for(int i=2;i<=n;i++)
        {
            if(!sign[i]&&isprime[i+a[m-1]])
            {
                a[m]=i;
                sign[i]=1;//已经用过，标记为1
                dfs(m+1);//递归
                sign[i]=0;//如果递归回来没有用到这个数，重新标记为可用的
            }
        }
    }
}
int main()
{
    int count=0;
    a[0]=1;
    while(~scanf("%d",&n))
    {
        memset(sign,0,sizeof(sign));
        count++;
        cout << "Case " << count <<":" << endl;
        dfs(1);
        cout << endl;
    }
    return 0;
}