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

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 using namespace std;
 5 int main()
 6 {
 7     int n,j,i,k,a[1005],v[1005];
 8     while(~scanf("%d",&n))
 9     {
10         for(i=1;i<=n;i++)
11           scanf("%d",&a[i]);
12        memset(v,0,sizeof(v));
13        int cnt=0;
14        int flag=0;
15        while(cnt<n)
16        {
17            for(i=1;i<=n;i++)
18            {
19                if(v[i]==1)
20                   continue;
21                   k=0;
22                 for(j=i+1;j<=n&&k<=5;j++)
23                 {
24                     if(v[j]==1)
25                       continue;
26                     if(a[i]==a[j])
27                     {
28                        v[i]=v[j]=1;
29                         break;
30                     }
31                     k++;
32                 }
33            }
34            cnt++;
35        }
36        for(i=1;i<=n;i++)
37         {
38                if(v[i]==0)
39                {
40                    printf("0\n");
41                    break;
42                }
43         }
44         if(i>n)
45           printf("1\n");
46
47     }
48     return 0;
49 }

HDU-4272 LianLianKan