HDU4272LianLianKan（dfs）

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

 

 

Source

2012 ACM/ICPC Asia Regional Changchun Online

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注意给的顺序是从底部到顶部。

开始想直接模拟，不可行，因为在下面5个里可能有一个或多个相同的元素，每个选择都可能对后面造成影响，因此要dfs。

其中还巧妙的用了map，当然改用set来存然后看里面到底有没有奇数个的元素也是可以的，这点的优化很重要。

 #include<stdio.h>
 #include<iostream>
 #include<string>
 #include<cstring>
 #include<algorithm>
 #include<queue>
 #include<set>
 #include<map>
 
 using namespace std;
 
 int vis[],a[];
 
 int dfs(int n)
 {
     int i=,j=n-;
     while(n>=&&vis[n]) --n;
     if(n==-) return ;
     if(n==&&!vis[]) return ;
     while(i<=)
     {
         if(j<) return ;
         if(vis[j]) --j;
         if(a[n]==a[j])
         {
             vis[j]=;
             if(dfs(n-)) return ;
             vis[j]=;
         }
         ++i;
         --j;
     }
     return ;
 }
 
 int main()
 {
     int k,m,q,t,p,n;
     int T;
     map<int,int> mp;
     map<int,int>::iterator it;
 
     while(cin>>n)
     {
         t=;
         for(int i=;i<n;++i)
         {
             scanf("%d",&a[i]);
             vis[i]=;
             mp[a[i]]++;
         }
 
         for(it=mp.begin();it!=mp.end();++it)
         {
             if((*it).second%)
             {
                 t=;
                 break;
             }
         }
 
         if(t)
         {
             cout<<<<endl;
         }else
         {
             cout<<dfs(n-)<<endl;
         }
 
     }
     return ;
 }