HDU 1016 Prime Ring Problem（dfs）

最新推荐文章于 2019-05-22 21:04:01 发布

柏油

最新推荐文章于 2019-05-22 21:04:01 发布

阅读量211

点赞数 1

CC 4.0 BY-SA版权

分类专栏： # 图论 DFS

本文链接：https://blog.youkuaiyun.com/ldw201510803006/article/details/62455809

图论 DFS 专栏收录该内容

40 篇文章

订阅专栏

本文介绍了一种使用深度优先搜索（DFS）算法解决素数环问题的方法。该问题要求在一个由n个节点组成的环中放置1到n的自然数，使得任意相邻两节点上的数字之和为素数。代码实现中包含了素数判断及输出所有符合条件的环状序列。

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 47896 Accepted Submission(s): 21158

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：dfs求解，相邻两个数素数的判定（当前数加上之前一个数进行判定，第一个数和最后一个数进行判定）

AC代码：

#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int maxn=20+1;
int vis[maxn];
int a[maxn];
int n;

bool is_prime(int i1,int i2){
	for(int j=2;j<=sqrt(i1+i2);j++)
	if((i2+i1)%j==0)return false ;	
	return true;
}
void dfs(int cur){
	if(cur>n && is_prime(a[1],a[n])){
		printf("%d",a[1]);
		for(int i=2;i<=n;i++)
		printf(" %d",a[i]);
		printf("\n");
	}
	for(int i=1;i<=n;i++){
		if(!vis[i]){
			if(is_prime(a[cur-1],i)){
			a[cur]=i;
			vis[i]=1;
			dfs(cur+1);
			vis[i]=0;
			}
		}
	}
}
int main(){
	int cnt=0;
	while(scanf("%d",&n)==1){
	  memset(vis,0,sizeof(vis));
	  a[1]=1;
	  vis[1]=1;
	  printf("Case %d:\n",++cnt);
	  dfs(2);
	  printf("\n");
	  	}
	return 0;
}