hdu 1016 Prime Ring Problem dfs

来源：互联网发布：js清空数组中所有元素编辑：程序博客网时间：2024/04/27 07:32

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36533 Accepted Submission(s): 16102

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

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      题目意思：  给你一个整数 n  求出所有由 数字1 2 3 4 5 6..... n 中锁有数字组成的素数环并输出出。                素数环的定义： 环中任意相邻的两个数的和是一个素数；          思路：  直接dfs 搜一遍就可以了，  n最大是20 不会超时      我的代码：#include<bits/stdc++.h>using namespace std;const int maxn = 22;int vis[maxn];int n;int ans[maxn];int cheak(int k)          //函数功能是 检查k是不是素数{    int falt = 1;    for(int i = 2; i < k; i++)        if(k % i == 0)           falt = 0;    return falt;}int dfs(int t){    if(t == n +1 && cheak(ans[1] + ans[n]))   //如果检查成功了就输出环    {        for(int i = 1 ; i <= n; i++)             i == n ? printf("%d\n", ans[i]) : printf("%d ", ans[i]);    }    for(int i = 2; i <= n; i++)               //位置i的所有可能    {        if(cheak(i + ans[t - 1])&& !vis[i])    //检查        {            vis[i] = 1;                        //DFS  这里注意设为1， 为了避免重复及死递归            ans[t] = i;            dfs(t+1);            vis[i] = 0;                        //DFS  这里注意归0， 为了避免重复及死递归        }    }}int main(){   int Case = 1;   while(scanf("%d", &n) == 1 && n)   {       memset(vis, 0, sizeof(vis));  //将访问数组归0；       ans[1] = 1;                   // 1 被固定， 故在开始就设为1       vis[1] = 1;                   //1 被固定， 故在开始就设为1       printf("Case %d:\n", Case++);        dfs(2);                       // 从2 开始扫, 因为1被固定了       printf("\n");   }}

0 0