HDU/HDOJ 1016 简单dfs

最新推荐文章于 2018-07-26 19:11:13 发布

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最新推荐文章于 2018-07-26 19:11:13 发布

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本文链接：https://blog.youkuaiyun.com/WuBaizhe/article/details/54615762

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针对 PrimeRing 问题，本文详细介绍了如何使用深度优先搜索（DFS）结合有效的剪枝策略来解决这一经典问题。通过筛选素数和限制无效的搜索路径，有效提升了算法效率。

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Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 46409 Accepted Submission(s): 20500

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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题意很简单，dfs思路也很直接，但仍然TLE了两发，下面讲解剪枝思路：

1.因为题目只要求是以“1”开头，所以我们可以直接把1填入结果数组ans里面，从初始深度为1开始搜索。

2.因为任意两个相邻数和为质数，所以枚举时可以加个判断，若当前枚举数与结果数组前一个的和不为素数，则直接枚举下一个。

3.为了进一步减少时间，所以可以加个素数筛得到素数。

#include <bits/stdc++.h>
using namespace std;

int vis[25],ans[25],n,dp[1001];

void init(void){
	memset(dp,0,sizeof(dp));
	int i,j;
	dp[0] = dp[1] = 1;
	for(i=2 ;i<1001 ;i++){
		if(!dp[i]){
			for(j=2 ;i*j<1001 ;j++){
				dp[i*j] = 1;
			}
		}
	}
}

bool check(int* a){
	int i;
	if(dp[a[0] + a[n-1]])
		return false;
	return true;
}

void output(int* a){
	int i;
	printf("%d",a[0]);
	for(i=1 ;i<n ;i++){
		printf(" %d",a[i]);
	}
	printf("\n");
}

void dfs(int deepth){
	int i;
	if(deepth == n){
		if(check(ans)){
			output(ans);	
		}
		return;
	}
	for(i=2 ;i<=n ;i++){
		if(!vis[i]){
			if(dp[i+ans[deepth-1]])
				continue;
			ans[deepth] = i;
			vis[i] = 1;
			dfs(deepth+1);
			vis[i] = 0;
		}
	}
}

int main(){
	int _=1;
	init();
	while(scanf("%d",&n) != EOF){
		memset(vis,0,sizeof(vis));
		memset(ans,0,sizeof(ans));
		ans[0] = 1;
		printf("Case %d:\n",_++);
		dfs(1);
		printf("\n");
	}
	return 0;
}