poj Anti-prime Sequences

Anti-prime Sequences

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 1820		Accepted: 844

Description

Given a sequence of consecutive integers n,n+1,n+2,...,m, an anti-prime sequence is a rearrangement of these integers so that each adjacent pair of integers sums to a composite (non-prime) number. For example, if n = 1 and m = 10, one such anti-prime sequence is 1,3,5,4,2,6,9,7,8,10. This is also the lexicographically first such sequence.

We can extend the definition by defining a degree danti-prime sequence as one where all consecutive subsequences of length 2,3,...,d sum to a composite number. The sequence above is a degree 2 anti-prime sequence, but not a degree 3, since the subsequence 5, 4, 2 sums to 11. The lexicographically .rst degree 3 anti-prime sequence for these numbers is 1,3,5,4,6,2,10,8,7,9.

Input

Input will consist of multiple input sets. Each set will consist of three integers, n, m, and d on a single line. The values of n, m and d will satisfy 1 <= n < m <= 1000, and 2 <= d <= 10. The line 0 0 0 will indicate end of input and should not be processed.

Output

For each input set, output a single line consisting of a comma-separated list of integers forming a degree danti-prime sequence (do not insert any spaces and do not split the output over multiple lines). In the case where more than one anti-prime sequence exists, print the lexicographically first one (i.e., output the one with the lowest first value; in case of a tie, the lowest second value, etc.). In the case where no anti-prime sequence exists, output

No anti-prime sequence exists.

Sample Input

1 10 2
1 10 3
1 10 5
40 60 7
0 0 0

Sample Output

1,3,5,4,2,6,9,7,8,10
1,3,5,4,6,2,10,8,7,9
No anti-prime sequence exists.
40,41,43,42,44,46,45,47,48,50,55,53,52,60,56,49,51,59,58,57,54

开始看到这题想到深搜+回溯，看到数据范围1~1000没敢想，估计肯定会超时，之后自己想了想解题过程，好像还是没有思路，在网上搜了搜，竟然是深搜+回溯，还是自己对深搜的时间复查度不知道呀；
算法导论里dfs（）的实际那复查度O(v+e)
贴下自己的超时代码，以后再检查：

#include <cstdlib>
#include <iostream>

using namespace std;

int n,m,d,f[1010];
bool vis[1010];
bool isprime[1010];
bool find_prime()
{
     memset(isprime,1,sizeof(isprime));
     isprime[0]=isprime[1]=0;
     for(int i=2;i<1005;i++)
       if(isprime[i])
       {
          for(int j=2*i;j<1005;j+=i)
          {
             isprime[j]=0;        
          }              
       }
}

bool check(int cur,int n)
{
     if(cur<0)  return 1;
     int k;
    if(cur-d+1<0) k=0;
    else  k=cur-d+1;
    int sum=n;
    for(int i=cur-1;i>=k;i--)
    {
        sum+=f[i];
        if(isprime[sum])  return 0;        
    }
 //   if(isprime(sum))  return 0;
    return 1;     
}

bool dfs(int cur)
{
    if(cur>=n-m+1)  return 1;
    for(int i=m;i<=n;i++)
    {
        if(!vis[i]&&check(cur,i)) 
        {
           vis[i]=1;
           f[cur]=i;
           if(dfs(cur+1))
             return 1;
           vis[i]=0;                            
        }       
    } 
    return 0;    
}

int main(int argc, char *argv[])
{
    find_prime();
    while(scanf("%d%d%d",&m,&n,&d))
    {
         if(m==0&&n==0&&d==0)  break;                          
         for(int i=m;i<=n;i++)
           vis[i]=0;                                   
         if(dfs(0))
         {
             for(int i=0;i<n-m;i++)
              printf("%d,",f[i]);          
              printf("%d\n",f[n-m]);           
         }                                   
         else
           puts("No anti-prime sequence exists.");                                       
    }
    system("PAUSE");
    return EXIT_SUCCESS;
}