HDUOJ1016 Prime Ring Problem

 

Prime Ring Problem

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

Problem DescriptionA 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.

 

 

Inputn (0 < n < 20). 

 

OutputThe 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

68

 

 

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2【解題思路】素數環是典型的DFS題目，DFS函數的參數為(int c, int cnt)，c代表遍曆時所在環上節點的數，cnt代表結點個數。當節點個數位n時並且環末節點和首節點相加也為素數則列印環。
根據本題總結dfs的主要步驟和方法：
1． 找搜尋出口條件，即搜尋到最深層時的操作
2．  Dfs採用遞迴搜尋，在遍曆後返回時標記已經遍曆的節點
3． 注意允入準則的滿足
4． 畫出遞迴搜尋圖查看函數是否滿足要求
5.  注意回溯的條件，適當的時候需要剪枝
 
 
 
//============================================================================// 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;}