HDU 1016 Prime Ring Problem（经典DFS+回溯）

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原文链接：http://www.cnblogs.com/Blackops/p/5766278.html

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Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42600 Accepted Submission(s): 18885

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)

题目链接：HDU 1016

经典的DFS回溯题目，以前一直想做来着，但是不懂回溯搜索，现在类似的一些题还是挺简单的……，这题用输出外挂可以优化到200+MS，题意是把1-n中所有自然数全部排完才能算一个环，刚开始搞错了输出爆炸……

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define MM(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=50;
int prime[N];
int pos[N],vis[N];
int n;
void Out(int a)
{
    if(a>9)
        Out(a/10);
    putchar(a%10+'0');
}
inline bool check()
{
	for (int i=1; i<=n; ++i)
	{
		if(!vis[i])
			return false;
	}
	return true;
}
void dfs(int now)
{
	if(now==n)
	{
		if(prime[pos[now]+pos[1]]&&check())
		{
			for (int i=1; i<=n; ++i)
			{
				Out(pos[i]);
				putchar(i==n?'\n':' ');
			}
		}
		return ;
	}
	for (int i=2; i<=n; ++i)
	{
		if(!vis[i]&&prime[i+pos[now]]&&now<=n)
		{
			vis[i]=1;
			pos[now+1]=i;
			dfs(now+1);
			pos[now+1]=0;
			vis[i]=0;
		}
	}
}
int main(void)
{
	int i,j;
	for (i=0; i<N; ++i)
		prime[i]=1;
	prime[1]=0;
	for (i=2; i<N; ++i)
		for (j=2; j*i<N; ++j)
			prime[i*j]=0;
	int tcase=0,m;
	while (~scanf("%d",&n))
	{
		MM(pos,0);
		vis[1]=1;
		pos[1]=1;
		printf("Case %d:\n",++tcase);
		dfs(1);
		putchar('\n');
	}
	return 0;
}

转载于:https://www.cnblogs.com/Blackops/p/5766278.html