HDU-1016 Prime Ring Problem (DFS)

    题目:HDU 1016

    题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1016

    题目：

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37741    Accepted Submission(s): 16682

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

   
   
   
   

    
    
    
    6
8
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case 1:
1 4 3 2 5 6
1 6 5 2 3 4

Case 2:
1 2 3 8 5 6 7 4
1 2 5 8 3 4 7 6
1 4 7 6 5 8 3 2
1 6 7 4 3 8 5 2
   
   
   
   

 

     这道题目是非常经典的dfs，题意是给一个n（n<20），用1~n构建素数环，使任意两个相邻整数之和为素数，所以直接用dfs来做，因为n取值范围是20，我们直接把40以内的12个素数打印出来，因为以1开始，旁边的数和1的和必须是素数而且和1不同，那2肯定不合适，所以就变成了11个素数（能省则省嘛~），之后就是深搜，如果满足素数换，那么就输出，不满足就回溯，我做这道题目时卡在了将状态回归，也就是在递归之后消除上次递归对下次递归影响这里，虽然卡了很久但最后搞懂之后还是挺有收获的，不多说废话，自己感觉一下就知道了。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<math.h>
using namespace std;
const int maxn=25;
int n;
int prime[11]={3,5,7,11,13,17,19,23,29,31,37};  //打表
int visit[maxn];                                //标记是否走过
int ans[maxn];                                  //存储答案（也就是序列）
int IsPrime(int x){                             //判断是否是素数，仅仅用于判断队尾和队首和是不是素数
    if(x==1) return 0;
    for(int i=2;i<=sqrt((double)x);i++)
        if(x%i==0) return 0;
        return 1;
}
void dfs(int x){                                //递归
    if(x==n) {                                  //满足条件输出答案
    if(IsPrime(ans[n-1]+ans[0])){
        for(int i=0;i<n-1;i++)
            printf("%d ",ans[i]);
            printf("%d\n",ans[n-1]);
    }
    }
    else{
        for(int i=0;prime[i]-ans[x-1]<=n && x<n;i++){ //循环
            int temp=prime[i]-ans[x-1];
            if(temp>1 && temp<=n){
                if(visit[temp]==0){
                    ans[x]=temp;
                    visit[temp]=1;
                    dfs(x+1);
                    visit[temp]=0;                //消除标记影响
            }
        }
        }
    }
    return;
}
int main(){
    int count=0;
    while(cin>>n){
        printf("Case %d:\n",++count);
        memset(ans,0,sizeof(ans));
        memset(visit,0,sizeof(visit));
        ans[0]=1;
        visit[1]=1;
        dfs(1);
        printf("\n");
    }
    return 0;
}

     好好学习，天天向上~~~