hdu1016：Prime Ring Problem

原题链接

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 47623    Accepted Submission(s): 21048

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)

 

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题意： 

给你一个整数n， 求解1~n排列成环使得每相邻的两个数相加是素数的所有情况。

解题思路： DFS

DFS从x = 1开始深搜， 每次找到符合条件的两个数就加入数组a，当满足x == n 并且a[x] + a[1] 也是素数时结束， 并输出a数组。

代码：

#include <cmath>#include <string>#include <cstdio>#include <sstream>#include <cstring>#include <iostream>#include <algorithm>#define LOCALusing namespace std;int n;int a[105];int visit[105];int is_prime[105];void f()              //素数打表， 也可直接写入前40个素数入该数组{    for (int i = 2; i <= 10; i++)        if(!is_prime[i])            for (int j = i+i; j <= 100; j+=i)                is_prime[j] = 1;}void dfs(int x){    if(x == n && !is_prime[a[x]+a[1]])  //满足条件并输出    {        for (int i = 1; i < n; i++)            printf("%d ",a[i]);        printf("%d\n", a[x]);    }    else    {        for (int i = 2; i <= n; i++)        {            if(!visit[i])                   //判断是否访问过            {                if(!is_prime[i + a[x]])     //判断i是否符合当前条件                {                    visit[i] = 1;       //标记已访问                    a[++x] = i;         //i加入a数组                    dfs(x);                    x--;                //退一格                    visit[i] = 0;       //取消标记，以便之后DFS                }            }        }    }}int main(){//    #ifdef LOCAL//    freopen("data.in", "r", stdin);//    freopen("data.out", "w", stdout);//    #endif    f();    int z = 1;    while (scanf("%d", &n) != EOF)    {        memset(visit, 0, sizeof(visit));        a[1] = 1;                   //初始时从1开始        visit[1] = 1;               //标记1已经走过，虽然并不会影响。        printf("Case %d:\n", z++);        dfs(1);        printf("\n");    }    return 0;}