HDU-4272 LianLianKan

          http://acm.hdu.edu.cn/showproblem.php?pid=4272

        据说是状态压缩，+dfs什么什么的，可我这样也过了，什么算法都是浮云 ，暴力才是王道。我也归类为状态压缩，可以用状态压缩来做。

           LianLianKan

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2482    Accepted Submission(s): 773

Problem Description

I like playing game with my friend, although sometimes looks pretty naive. Today I invent a new game called LianLianKan. The game is about playing on a number stack.
Now we have a number stack, and we should link and pop the same element pairs from top to bottom. Each time, you can just link the top element with one same-value element. After pop them from stack, all left elements will fall down. Although the game seems to be interesting, it's really naive indeed.

To prove I am a wisdom among my friend, I add an additional rule to the game: for each top element, it can just link with the same-value element whose distance is less than 6 with it.
Before the game, I want to check whether I have a solution to pop all elements in the stack.

 

 

Input

There are multiple test cases.
The first line is an integer N indicating the number of elements in the stack initially. (1 <= N <= 1000)
The next line contains N integer ai indicating the elements from bottom to top. (0 <= ai <= 2,000,000,000)

 

 

Output

For each test case, output “1” if I can pop all elements; otherwise output “0”.

 

 

Sample Input

2

1 1

3

1 1 1

2

1000000 1

 

 

Sample Output

1

0

0

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
    int n,j,i,k,a[1005],v[1005];
    while(~scanf("%d",&n))
    {
        for(i=1;i<=n;i++)
          scanf("%d",&a[i]);
       memset(v,0,sizeof(v));
       int cnt=0;
       int flag=0;
       while(cnt<n)
       {
           for(i=1;i<=n;i++)
           {
               if(v[i]==1)
                  continue;
                  k=0;
                for(j=i+1;j<=n&&k<=5;j++)
                {
                    if(v[j]==1)
                      continue;
                    if(a[i]==a[j])
                    {
                       v[i]=v[j]=1;
                        break;
                    }
                    k++;
                }
           }
           cnt++;
       }
       for(i=1;i<=n;i++)
        {
               if(v[i]==0)
               {
                   printf("0
");
                   break;
               }
        }
        if(i>n)
          printf("1
");

    }
    return 0;
}