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

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

     }
     return ;
 }