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= 25;
const int maxm= 1e4+10;
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= 2;i <=tmp; i++)
        if(num %i== 0)
          return 0 ;
     return 1 ;
}
void dfs(int x)
{
    visit[x]=1;    //标记访问过
    order[cnt++]=x; //记录路径
    for(int i=0;i<v[x].size();i++)    //df搜边
    {
        if(visit[v[x][i]]==0)
        {
            dfs(v[x][i]);
        }
    }
    visit[x]=0;    //回溯
    if(cnt==n&&isPrime(x+1))   //长度为n且最后一个数与1相加为素数
    {
        for(int i=0;i<cnt;i++) 
        {
            if(i==cnt-1)
                printf("%d
",order[i]);
            else
                printf("%d ",order[i]);
        }
    }
    cnt--;
}
int main()
{
    int kase=1;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
            v[i].clear();
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(isPrime(i+j)==1&&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,0,sizeof(visit));   //初始化访问数组 0未访问过
        printf("Case %d:
",kase++);
        cnt=0;   //路径长度初始化
        dfs(1);  
        printf("
");
    }
}