#HDU 1016 Prime Ring Problem 【DFS+溯回求组数】

题目：

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38251    Accepted Submission(s): 16926

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

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

 

Source

Asia 1996, Shanghai (Mainland China)

 

Recommend

JGShining

题意即，给定数字N，将1~N共N个数字排序，使得相邻两数之和为质数，求出所有符合的排序。

首先我们要求出20以内所有质数制成质数表，方便查询。

对于每次搜索，初始值一定为1（题目规定），第二位有（N-1）种选择，即从初始状态可DFS递推至（N-1）条分路。

对于每一条路，又可以有（N-2）种选择。为了减少复杂度，当我们发现待选的路径和上一位不满足加和为质数时，我们将对这条路进行剪枝。即直接break掉，溯回上一层重新选择。

最后输出所有可能即可。

#include<iostream>#include<algorithm>#include<string.h>using namespace std;int data[25][25];int ans[25];int que[25], n, be;int sim[20] = { 0, 2, 3, 5, 7, 11, 13, 17, 19 };int finding(int a, int b){for (size_t i = 2; i <= 8; i++){if ((a + b) % i == 0 && (a + b) != i)//建立质数表{return 1;}}return 0;}void fins(){for (size_t s = 1; s < 21; s++){for (size_t i = 1; i < 21; i++){if (!finding(s, i))//按层进行搜索{data[s][i] = 1;}}}}void print(){for (size_t i = 0; i < n - 1; i++){cout << que[i] << " ";//que数组用来存已选择的数字，确定满足条件后 按位输出}cout << que[n - 1];cout << "\n";}void dfs(int s){for (size_t i = 1; i <= n; i++){if (data[s][i] == 1 && ans[i] == 0){ans[i] = 1;que[be] = i;be++;if (be == n){if (data[i][1] == 1){print();}}else{dfs(i);}be--;ans[i] = 0;}}}int main(){int cas = 1;fins();while (cin >> n){cout << "Case " << cas << ":\n";be = 0;memset(ans, 0, sizeof(ans));memset(que, 0, sizeof(que));ans[1] = 1;que[be] = 1;be++;dfs(1);cout << "\n";cas++;}}