HDU1016 DFS+回溯（保存路径）

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 55053    Accepted Submission(s):
24366

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

题意   给你一个n  对1 ... n   n个数排序 1必须在第一位 两个数相加要为素数（1与最后一个数相加也必须为素数） 输出所有满足上述的序列。

解析 我们先1~n分别求出与它相加为素数的数存进数组，然后就相当于建了一个图 从1出发 一层一层向下推 基础DFS 然后在回溯一下 记录路径就好了

AC代码

 #include <stdio.h>
 #include <math.h>
 #include <string.h>
 #include <stdlib.h>
 #include <iostream>
 #include <sstream>
 #include <algorithm>
 #include <string>
 #include <queue>
 #include <vector>
 using namespace std;
 const int maxn= ;
 const int maxm= 1e4+;
 const int inf = 0x3f3f3f3f;
 typedef long long ll;
 int visit[maxn],order[maxn];
 vector<int> v[maxn];
 int n,cnt;
 int isPrime(int num )
 {
      int tmp =sqrt( num);
      for(int i= ;i <=tmp; i++)
         if(num %i== )
           return  ;
      return  ;
 }
 void dfs(int x)
 {
     visit[x]=;    //标记访问过
     order[cnt++]=x; //记录路径
     for(int i=;i<v[x].size();i++)    //df搜边
     {
         if(visit[v[x][i]]==)
         {
             dfs(v[x][i]);
         }
     }
     visit[x]=;    //回溯
     if(cnt==n&&isPrime(x+))   //长度为n且最后一个数与1相加为素数
     {
         for(int i=;i<cnt;i++)
         {
             if(i==cnt-)
                 printf("%d\n",order[i]);
             else
                 printf("%d ",order[i]);
         }
     }
     cnt--;
 }
 int main()
 {
     int kase=;
     while(scanf("%d",&n)!=EOF)
     {
         for(int i=;i<=n;i++)
             v[i].clear();
         for(int i=;i<=n;i++)
         {
             for(int j=;j<=n;j++)
             {
                 if(isPrime(i+j)==&&i!=j)    //保存能够与i相加为素数的数
                     v[i].push_back(j);
             }
         }
 //        for(int i=1;i<=n;i++)
 //        {
 //            cout<<i;
 //            for(int j=0;j<v[i].size();j++)
 //                printf(" %d",v[i][j]);
 //            cout<<endl;
 //        }
         memset(visit,,sizeof(visit));   //初始化访问数组 0未访问过
         printf("Case %d:\n",kase++);
         cnt=;   //路径长度初始化
         dfs();
         printf("\n");
     }
 }