杭电——1016 Prime Ring Problem 深搜

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Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 48854    Accepted Submission(s): 21541

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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PS;本人写的第一道深搜题目，觉得我写的不好的大神可以选择不看。

题意：输入一个偶数n，然后找出1~n的组合形成一个环，使得相邻两个数之和为素数，即形成素数环；输出所有符合条件的组合素数环。输入数据有多组，其中1<n<20.

解题思路：

1.由于所求数据在20以内，相加之后不会超过40，所有可以提前预处理一下做个40以内的素数判断数组，以节约用时。

2.接着就是做深搜循环判断了。

代码如下：

#include <string.h>#include <stdio.h>#define M 40int vis[M];                //标记数组int dis[M];               //素数环数组int a[M];//素数判断数组int n;void sushu()     //素数预处理，若数组a的下标为素数，那么a[i]=1，否则a[i]=0，（本人还是习惯用1，0表示，不习惯用bool类型,但推荐用bool型，我也会慢慢改变的）{    int i,j;    memset(a,1,sizeof(a));    for(i=2;i<M;i++)        {            if(a[i])                for(j=i+i;j<M;j+=i)          //是i的倍数都不是素数                a[j]=0;                     //不是素数的标记为0        }}void dfs(int x)                          //x表示已判断个数{    int i;    if(x==n&&a[dis[n-1]+dis[0]])          //当判断了n个数并且dis[n-1]+dis[0]也是素数的的时候输出该素数环     {         for(i=0;i<n-1;i++)        printf("%d ",dis[i]);        printf("%d\n",dis[n-1]);    }    else    {        for(i=2;i<=n;i++)            if(vis[i]&&a[dis[x-1]+i])        {            vis[i]=0;           //符合条件，标记为0，表示该数已被使用，不可再用            dis[x]=i;            dfs(x+1);            vis[i]=1;//回溯        }    }}int main(){    int t=1;    sushu();    while(~scanf("%d",&n))    {        memset(vis,1,sizeof(vis));      //记得每次初始化数组        printf("Case %d:\n",t++);        vis[1]=0;        dis[0]=1;        dfs(1);        printf("\n");    }    return 0;}