HDOJ 1016 Prime Ring Problem（dfs+回溯）

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33376    Accepted Submission(s): 14770

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

 

ac代码：

#include<stdio.h>#include<string.h>int n;int num[20];int v[20];int check(int x)//判断素数{    for(int i=2;i<x;i++)    if(x%i==0)    return 0;    return 1;}void dfs(int s){    int i,j;    if(s==n+1&&check(num[n]+num[1]))//最后一个数和第一个数的判断    {        for(i=1;i<=n-1;i++)        {            printf("%d ",num[i]);        }        printf("%d\n",num[n]);//这题坑了两下，杭电输出真心严        return;    }    for(j=2;j<=n;j++)    {        if(check(j+num[s-1])&&v[j]==0)        {            v[j]=1;            num[s]=j;            dfs(s+1);            v[j]=0;//回溯        }    }}int main(){    int i,m=1;    while(scanf("%d",&n)!=EOF)    {        memset(v,0,sizeof(v));        num[1]=1;        printf("Case %d:\n",m++);        dfs(2);        printf("\n");    }    return 0;}