hdoj 4272 LianLianKan 数据太水

LianLianKan

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2884    Accepted Submission(s): 898

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

 

 

题意

给你一个队列，然后每个子可以和距离为以内的子连接，然后这两个子都pop,然后问你，是否这个队列能够全部pop掉

 

 

题解

数据太水了，判一判子的个数是否为偶数，然后再判断是否每一类都是偶数，然后直接输出1，然后就A了= =

标答应该是一个状态压缩DP

 

LL a[maxn];
int vis[maxn];
int flag=0;
map<int,int> kiss;
int n;
void dfs(int nn,int cur)
{
    if(nn>n)
        return;
    if(vis[nn]==1)
        dfs(nn+1,cur);
    if(cur==n)
    {
        flag=1;
        return;
    }
    int m=nn;
    REP_1(i,5)
    {
        m++;
        while(1)
        {
            if(m>n)
                return;
            if(vis[m]==0)
                break;
            m++;
        }
        if(a[m]==a[nn])
        {
            vis[nn]=1;
            vis[m]=1;
            dfs(nn+1,cur+2);
            vis[nn]=0;
            vis[m]=0;
        }
    }
}
int main()
{
    while(RD(n)!=-1)
    {
        memset(vis,0,sizeof(vis));
        kiss.clear();
        REP_1(i,n)
        {
            RD(a[i]);
            kiss[a[i]]++;
        }
        if(n%2==1)
        {
            flag=0;
            printf("0\n");
            continue;
        }
        int flag2=1;
        REP_1(i,n)
        {
            if(kiss[a[i]]%2==1)
            {
                flag2=0;
                break;
            }
        }
        if(flag2==0)
        {
            printf("0\n");
            continue;
        }
        //dfs(1,0);
        cout<<"1"<<endl;
    }

}