HDUOJ1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8087Accepted Submission(s): 3620

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
【解题思路】素数环是典型的DFS题目，DFS函数的参数为(int c, int cnt)，
c代表遍历时所在环上节点的数，cnt代表结点个数。当节点个数位n时并且环末节点
和首节点相加也为素数则打印环。
根据本题总结dfs的主要步骤和方法：
1． 找搜索出口条件，即搜索到最深层时的操作
2． Dfs采用递归搜索，在遍历后返回时标记已经遍历的节点
3． 注意入口条件的满足
4． 画出递归搜索图查看函数是否满足要求
5.  注意回溯的条件，适当的时候需要剪枝
//============================================================================
// Name        : HDUOJ1016.cpp
// Author      : Weisi Shi
// Version     :
// Copyright   : Weisi Shi
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
using namespace std;
int n;
bool p[41];
int circle[21];
bool visit[21];
void prime() {
	int i, j;
	memset(p, true, sizeof(p));
	for (i = 2; i < 41; ++i) {
		for (j = 2; i * j < 41; ++j) {
			p[i * j] = false;
		}
	}
}
void dfs(int c, int cnt) {
	int i, j;
	if (cnt == n && p[circle[1] + circle[n]]) {
		for (i = 1; i < n; ++i)
			cout << circle[i] << " ";
		cout << circle[n] << endl;
	}
	for (j = 1, i = c + j; j <= n; i++, j++) {
		if (p[i] && !visit[j]) {
			circle[cnt + 1] = j;
			visit[j] = true;
			dfs(j, cnt + 1);
			visit[j] = false;//回溯
		}
	}
}
int main() {
	int c = 1;
	prime();
	while (cin >> n) {
		memset(visit, false, sizeof(visit));
		cout << "Case " << c++ << ":" << endl;
		visit[1] = true;
		circle[1] = 1;
		dfs(1, 1);
		cout << endl;
	}
	return 0;
}